0
comments

Wednesday, August 29, 2012

The Maria Theresa Thaler 1780

The Maria Theresa Thaler is probably one of the most famous and well known coins of the world.
Originally struck in Austria from 1740 to 1780, the Thaler was the currency of the Austrian Empire. It was very important for trade with the Levant (parts of Turkey, Lebanon, Syria). Over time, the Maria Theresa Thaler became the best known and most popular silver coin in the Arabian world. After the death of Empress Maria Theresa in 1780, Joseph II permitted the Austrian mint to continue striking the coin with the 1780 dies in order to meet demand from the Middle East. The 1780 taler was the only silver coin that the Arabs trusted and would accept. Since then, the Maria Theresa Thaler has been restruck for trade purposes at Vienna, Austria with the 1780 date frozen in time. The taler became the unofficial currency in some areas of Africa and Asia, and may still be in use today as a "trade silver dollar" in some Arabian bazaars.
Particularly on older strikes earlier than about 1850, almost each individual coin shows minor differences. Such differences help specialists to determine the origin and strike date of individual specimen. A close examination of more than 20 coins struck in Venice between 1817 and 1833 showed that nearly all coins were struck from different dies. Major characteristics of common variants are described in the List of variants.
Various articles claim that the Maria Theresa Thaler was re-struck with an unchanged appearance since 1780. However, this is only really true for restrikes made after about 1850. Earlier strikes are usually relatively easy to identify and classify. Unfortunately, the statement may cause collectors - and sometimes dealers - to believe that their coins might be original coins struck in 1780 or shortly thereafter. Sometimes such coins will actually be sold as "original strike". In reality, most of those "originals" will be post-1850 strikes.
In Original and Restrike we compare two coins struck in 1780 (Vienna mint) and a coin struck around 1781 (Guenzburg mint) with a modern restrike. This may help to identify the differences.
The Talers provides a list of major variants. This list is not complete and is being extended on an ongoing basis. There are somewhere between 100 and 150 major variants, and an uncountable number of strikes with minor differences.

Maria Theresia Taler Variants

To describe Maria Theresia Taler variants, we follow the classification and numbering scheme introduced by Dr. Franz Leypold in his book Der Maria Theresia Taler 1780 .
Maria Theresia Talers have a number of distinct features which can be used to determine the mint and strike date of an individual specimen. The following table summarizes the major variants and approximate strike dates for Maria Theresia Talers struck with date 1780. 
  • Obverse types
  • Reverse types
  • Signatures
  • Saltires
  • Rarity: C=very common, 1-7: somewhat rare to extremely rare, with 6:up to 10 specimen known to exist; 7:less than 5 specimen known to exist.
  • There may always be deviations from the generic description. For example, a common characteristic of reverse type C is that the eagle has 1-3-1 tail feathers; however, variants with 1-2-1 tail feathers are known as well.

Original and Restrike

Various articles claim that the Maria Theresia Taler was re-struck with an unchanged appearance since 1780. However, this is only really true for restrikes made after about 1850. Earlier strikes are typically relatively easy to identify. Unfortunately, the statement may cause collectors - and sometimes dealers - to believe that their coins might be original coins struck in 1780 or shortly thereafter. Sometimes such coins will actually be sold as "original strike".
This statement is incorrect, even if considering strikes dated earlier than 1858 (i.e., while the Taler was an official currency) to be originals. Most of the so called "originals" have in reality been struck after 1858.
Below you will find two coins struck in 1780 (Vienna mint), a coin struck around 1781 (Guenzburg mint), and a current restrike. This may help to identify basic differences between earlier strikes and more recent restrikes.

Strike
Obverse
Revers

Vienna Mint, 1780

First (older) Variant
Approximately 7-9 specimen known worldwide.
Obverse H6Reverse H6

Vienna Mint, 1780

Second (newer) Variant
Approximately 5 specimen known worldwide.
Obverse H7aReverse H7a

Guenzburg Mint, around 1781

Only a single specimen known worldwide.
Assumption for the 1780 Guenzburg strike is that the letter A should be pointed instead of flat as in this specimen. Signature may be S.F. or S:F:.
Obverse H27vReverse H27v

Modern Restrike


Struck more than 300.000.000 times worldwide.
Obverse Type IIIReverse Type C
Key identification characteristics for the 1780 strikes are in particular the form of the brooche in the veil on the obverse (no pearls, except for the later Vienna mint strike), the form of the letter 'U' in the reverse (AUST.DUX instead of AVST.DUX), the form of the cross next to the the year on the obverse, and the form of the ' 7 ' in the year. Older Vienna mint strikes (earlier than 1850) are signed I.C.-F.A. instead of S.F.

List of Maria Theresia Talers 1780

This is a list of known variants of Maria Theresia Talers. This list is by its very nature incomplete (with new variants still being discovered) and is being extended on an ongoing basis. There are between 100 and 150 major variants, and an uncountable number of strikes with minor differences.
The talers are ordered using the numbering scheme introduced in "Lexicon of the Maria Theresien Taler 1780" by Walter Hafner. The following viewing options are available.

0
comments

Monday, August 27, 2012

The Life of Neil Armstrong

It was July 20, 1969 when Neil Armstrong spoke what must be considered the most famous words of the 20th century, "one small step for man, one giant leap for mankind". This, of course, was the day that men from Earth first set foot on the Moon. It was the culmination of years of research and development, success and failure, and bitter competition from our feared rivals. And it was the words of a 38 year old Neil Armstrong that echo in the annals of history.

Early Life
Neil Armstrong was born August 5, 1930 on a farm in Wapakoneta, Ohio. But proceeded to move around the state for most of his young life, before his family finally settled back in Wapakoneta in 1944. There Neil held many jobs around town, especially at the local airport as he was always fascinated with aviation. After starting flying lessons at the age of 15, he was awarded a pilots license on his 16th birthday, before he had earned a driver's license.
Naturally interested in aviation, Armstrong decided to pursue a degree in Aeronautical Engineering from Purdue University under the Holloway Plan. This scholarship sent deserving recipients to University for the completion of their bachelors degree before committing to at least three years of Naval service. At the conclusion of their time in the Navy, they would return to school for two more years of advanced study. Armstrong later completed his master's degree in aerospace engineering at the University of Southern California in 1970.
Armstrong was called to Pensacola Naval Air Station in Florida in 1949 before he could complete his degree. There he earned his wings at 20 years of age, making him the youngest pilot in his squadron. He then flew 78 combat mission in Korea, earning three medals, including the Korean Service Medal. But Armstrong was sent home before the conclusion of the war and finished his Bachelors degree in 1955.

Testing New Boundaries

After completing his degree at Purdue, Armstrong decided to try his hand as a test pilot. After an initial application to the National Advisory Committee for Aeronautics (NACA) -- the prelude to the creation of NASA -- was turned down he took a post at Lewis Flight Propulsion Laboratory in Cleveland, Ohio. However, it was less than a year before Armstrong transfered to Edwards Air force Base (AFB) in California to work with the NACA's High Speed Flight Station.
During his tenure at Edwards Armstrong conducted test flights of over 50 types of experimental aircraft, logging 2,450 hours of air time. Among his accomplishments in these aircraft, Armstrong was able to achieve speeds of Mach 5.74 (4,000 mph or 6,615 km/h) and an altitude of 63,198 meters (207,500 feet), but in the X-15 aircraft.
Armstrong, being an engineer by training, had a technical efficiency to his flying that was the envy of most of his colleagues. However, he was criticized by some of the non-engineering pilots, including Chuck Yeager and Pete Knight, that observed that his flying technique was too mechanical. They argued that flying was, at least in part, feel. Something that didn't come naturally to the engineers, and it was this fact that sometimes got them into trouble.
While Armstrong was a comparatively successful test pilot, he was involved in several ariel incidents. One of the most famous events involving Armstrong occurred on May 21, 1962 when he was sent to investigate Delamar Lake as a potential emergency landing site, flying an F-104. After an unsuccessful landing damaged the radio and hydraulic system, Armstrong headed toward Nellis Air Force Base. When attempting to land at Nellis, the tail hook of the plane lowered due to the damaged hydraulic system and caught the arresting wire on the air field. The plane slid out of control down the runway, dragging the anchor chain along with it.
Upon radioing back to Edwards, Milt Thompson was dispatched in a F-104B to retrieve Armstrong. However, Milt had never flown this particular aircraft, and ended up blowing one of the tires during a hard landing. The runway was then closed for the second time that day to clear the landing path of debris. A third aircraft was sent to Nellis, piloted by Bill Dana. But Bill almost landed his T-33 Shooting Star long, prompting Nellis to send the pilots back to Edwards using ground transportation.

Crossing Into Space

In 1957 Armstrong was selected for the ironically named Man In Space Soonest (MISS) program. Then in September of 1963 he was selected as the first American civilian to fly in space. (Russia launched civilian Valentina Tereshkova into space in June of 1963, beating Armstrong to space.)
In 1966 Armstrong flew on the Gemini 8 mission which launched March 16. Serving as Command Pilot, the crew was to complete the first ever docking with another space craft, an unmanned Agena target vehicle. After 6.5 hours in orbit they were able to dock with the craft, but due to complications they were unable to complete what would have been the third ever "extra-vehicular activity" -- now referred to as a space-walk.
Armstrong also served as the CAPCOM -- typically the only person who directly communicates with the astronauts during missions to space -- for Gemini 11. However, it was not until the Apollo program began that Armstrong ventured into space.

The Apollo Program

Armstrong's first tour of service to the Apollo program came as the commander of the back-up crew of the Apollo 8 mission, though he had been originally scheduled to back-up the Apollo 9 mission. (Had he remained as the Apollo 9 back-up commander he would have been slated to Command Apollo 12, not Apollo 11.)
Initially, it was planned that Buzz Aldrin, the Lunar Module Pilot would be the first to set foot on the Moon. However, it was determined that because of the positions of the astronauts in the module, it would require Aldrin to physically crawl over Armstrong to reach the hatch. As such, it was decided that it would be easier for Armstrong to exit the module first upon landing.
Apollo 11 touched down on the surface of the Moon on July 20, 1969, at which point Armstrong declared, "Houston, Tranquility Base here. The Eagle has landed." A hugh sigh of relief was said to have been breathed throughout mission control, as it was thought that Armstrong had merely seconds of fuel remaining before the thrusters cut and the lander plummeted to the surface. Armstrong and Aldrin exchanged congratulations before quickly preparing the lander to launch off the surface in case of an emergency.

Man's Greatest Achievement

On July 20, 1969, Armstrong made his way down the ladder from the Lunar Lander and, upon reaching the bottom declared "I'm going to step off the LEM now." As his left boot made contact with the surface he then spoke the words that defined a generation, "That's one small step for man, one giant leap for mankind."
Interestingly, he meant to say "one small step for a man," referring to himself. Otherwise the phrase is actually contrary, since as stated man would imply mankind. Armstrong later was reported as saying that he hoped future quotations would include the "a" parenthetically. However, the phrase is still usually conveyed as he originally spoke it.
About 15 minutes after exiting the module, Aldrin joined him on the surface and they set to investigating the environment on the lunar surface. They also planted the American flag on the surface. But because of a malfunction of the bottom extender of the flag, the flag appeared to be waving. This, of course, would be impossible since there is no air on the Moon. It was supposed to be repaired on a later mission, but because the astronauts liked the way it looked, it was kept in the same condition.
The final task carried out by Armstrong was to leave behind a package of memorial items in remembrance of deceased Soviet cosmonauts Yuri Gagarin and Vladimir Komarov, and Apollo 1 astronauts Gus Grissom, Ed White and Roger Chaffee. All told, Armstrong and Aldrin spent 2.5 hours on the lunar surface. Each of the subsequent landings allowed for more and more time on the surface, culminating in 21 hours of extra-vehicular time performed by the Apollo 17 astronauts.
The astronauts then returned to Earth, splashing down in the Pacific Ocean on July 24, 1969. Armstrong was awarded the Presidential Medal of Freedom, the highest honor bestowed upon civilians, as well as a host of other medals from NASA and other countries.

Life After Space

After a short stint as an administrator with NASA and the Defense Advanced Research Projects Agency (DARPA), Armstrong accepted a teaching position at the University of Cincinnati with the department of Aerospace Engineering. He held this appointment until 1979. Armstrong also served on two investigation panels. The first was after the Apollo 13 incident, while the second came after the Challenger explosion.
Armstrong then lived a life outside the public eye. He stopped signing autographs more than a decade ago when he came to know that people were selling items baring his signature for thousands of dollars. There have also been issues with individuals selling forgeries. Armstrong  occasionally made public comments when asked about NASA and current policy. The most recent statement, in early 2010, was in staunch criticism of the President's plan to cancel the Constellation program.
He died on 25th of August, 2012 at the age of 82. Armstrong had heart surgery earlier that month to relieve blocked coronary arteries, he died from complications that followed the cardiovascular procedures.
Source: space.about.com
 
0
comments

Sunday, August 26, 2012

Penguins: Beautiful creatures

Penguins


Penguins are flightless birds that are highly adapted for the marine environment. They are excellent swimmers, and can dive to great depths, (emperor penguins can dive to over 500 m). Their shape enables extreme agility underwater; the feet and tail act as a rudder while the flippers act as propellers. They also have a waterproof coat of short, overlapping feathers and a well-developed layer of fat for insulation. Penguins feed on small fish and krill, catching these one at a time. In turn, penguins become food for other marine animals, namely leopard seals and killer whales. On land their main predators are skuas and sheathbills (carnivorous birds that take both eggs and chicks).
All penguins are restricted to the Southern Hemisphere, but the greatest concentrations are on Antarctic coasts and subantarctic islands. There are eighteen species of penguins, four of which live in Antarctica.  Adelie and emperor penguins breed on the shores of the Antarctic continent and nearby islands while chinstraps breed on islands around Antarctica and gentoos are found on islands ranging from the Antarctic to the subantarctic. Another four species (King, royal, rockhopper and macaroni) live on the subantarctic islands.
0
comments

Projectiles

A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles. An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible). And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

By definition, a projectile has a single force that acts upon it - the force of gravity. If there were any other force acting upon an object, then that object would not be a projectile. Thus, the free-body diagram of a projectile would show a single force acting downwards and labeled force of gravity (or simply Fgrav). Regardless of whether a projectile is moving downwards, upwards, upwards and rightwards, or downwards and leftwards, the free-body diagram of the projectile is still as depicted in the diagram at the right. By definition, a projectile is any object upon which the only force is gravity.


Projectile Motion and Inertia

Many students have difficulty with the concept that the only force acting upon an upward moving projectile is gravity. Their conception of motion prompts them to think that if an object is moving upward, then there must be an upward force. And if an object is moving upward and rightward, there must be both an upward and rightward force. Their belief is that forces cause motion; and if there is an upward motion then there must be an upward force. They reason, "How in the world can an object be moving upward if the only force acting upon it is gravity?" Such students do not believe in Newtonian physics (or at least do not believe strongly in Newtonian physics). Newton's laws suggest that forces are only required to cause an acceleration (not a motion). Recall from the Unit 2 that Newton's laws stood in direct opposition to the common misconception that a force is required to keep an object in motion. This idea is simply not true! A force is not required to keep an object in motion. A force is only required to maintain an acceleration. And in the case of a projectile that is moving upward, there is a downward force and a downward acceleration. That is, the object is moving upward and slowing down.
To further ponder this concept of the downward force and a downward acceleration for a projectile, consider a cannonball shot horizontally from a very high cliff at a high speed. And suppose for a moment that the gravity switch could be turned off such that the cannonball would travel in the absence of gravity? What would the motion of such a cannonball be like? How could its motion be described? According to Newton's first law of motion, such a cannonball would continue in motion in a straight line at constant speed. If not acted upon by an unbalanced force, "an object in motion will ...". This is Newton's law of inertia.


Now suppose that the gravity switch is turned on and that the cannonball is projected horizontally from the top of the same cliff. What effect will gravity have upon the motion of the cannonball? Will gravity affect the cannonball's horizontal motion? Will the cannonball travel a greater (or shorter) horizontal distance due to the influence of gravity? The answer to both of these questions is "No!" Gravity will act downwards upon the cannonball to affect its vertical motion. Gravity causes a vertical acceleration. The ball will drop vertically below its otherwise straight-line, inertial path. Gravity is the downward force upon a projectile that influences its vertical motion and causes the parabolic trajectory that is characteristic of projectiles.

 

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. The only force acting upon a projectile is gravity!


Characteristics of a Projectile's Trajectory

As discussed earlier in this lesson, a projectile is an object upon which the only force acting is gravity. Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward or downward they are also moving horizontally. There are the two components of the projectile's motion - horizontal and vertical motion. And since perpendicular components of motion are independent of each other, these two components of motion can (and must) be discussed separately. The goal of this part of the lesson is to discuss the horizontal and vertical components of a projectile's motion; specific attention will be given to the presence/absence of forces, accelerations, and velocity.

Horizontally Launched Projectiles

Let's return to our thought experiment from earlier in this lesson. Consider a cannonball projected horizontally by a cannon from the top of a very high cliff. In the absence of gravity, the cannonball would continue its horizontal motion at a constant velocity. This is consistent with the law of inertia. And furthermore, if merely dropped from rest in the presence of gravity, the cannonball would accelerate downward, gaining speed at a rate of 9.8 m/s every second. This is consistent with our conception of free-falling objects accelerating at a rate known as the acceleration of gravity.
If our thought experiment continues and we project the cannonball horizontally in the presence of gravity, then the cannonball would maintain the same horizontal motion as before - a constant horizontal velocity. Furthermore, the force of gravity will act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance as it did when it was merely dropped from rest (refer to diagram below). However, the presence of gravity does not affect the horizontal motion of the projectile. The force of gravity acts downward and is unable to alter the horizontal motion. There must be a horizontal force to cause a horizontal acceleration. (And we know that there is only a vertical force acting upon projectiles.) The vertical force acts perpendicular to the horizontal motion and will not affect it since perpendicular components of motion are independent of each other. Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.

The above information can be summarized by the following table.

Horizontal

Motion

Vertical

Motion

Forces

(Present? - Yes or No) (If present, what dir'n?)
No
Yes The force of gravity acts downward

Acceleration

(Present? - Yes or No) (If present, what dir'n?)
No
Yes "g" is downward at 9.8 m/s/s

Velocity

(Constant or Changing?)
Constant
Changing (by 9.8 m/s each second)
 

Non-Horizontally Launched Projectiles

Now suppose that our cannon is aimed upward and shot at an angle to the horizontal from the same cliff. In the absence of gravity (i.e., supposing that the gravity switch could be turned off) the projectile would again travel along a straight-line, inertial path. An object in motion would continue in motion at a constant speed in the same direction if there is no unbalanced force. This is the case for an object moving through space in the absence of gravity. However, if the gravity switch could be turned on such that the cannonball is truly a projectile, then the object would once more free-fall below this straight-line, inertial path. In fact, the projectile would travel with a parabolic trajectory. The downward force of gravity would act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance in every second as it did when it was merely dropped from rest (refer to diagram below). Once more, the presence of gravity does not affect the horizontal motion of the projectile. The projectile still moves the same horizontal distance in each second of travel as it did when the gravity switch was turned off. The force of gravity is a vertical force and does not affect horizontal motion; perpendicular components of motion are independent of each other.

 

In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory. This downward force and acceleration results in a downward displacement from the position that the object would be if there were no gravity. The force of gravity does not affect the horizontal component of motion; a projectile maintains a constant horizontal velocity since there are no horizontal forces acting upon it.
 


Check Your Understanding

Use your understanding of projectiles to answer the following questions. When finished, click the button to view your answers.
1. Consider these diagrams in answering the following questions.
Which diagram (if any) might represent ...
a. ... the initial horizontal velocity? b. ... the initial vertical velocity?
c. ... the horizontal acceleration?
d. ... the vertical acceleration?
e. ... the net force?
 

 



2. Supposing a snowmobile is equipped with a flare launcher that is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)?
a. in front of the snowmobile b. behind the snowmobile
c. in the snowmobile

 

 


 
3. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane?
a. below the plane and behind it. b. directly below the plane
c. below the plane and ahead of it

 

 

 
 So far  you have learned the following conceptual notions about projectiles.
  • A projectile is any object upon which the only force is gravity,
  • Projectiles travel with a parabolic trajectory due to the influence of gravity,
  • There are no horizontal forces acting upon projectiles and thus no horizontal acceleration,
  • The horizontal velocity of a projectile is constant (a never changing in value),
  • There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
  • The vertical velocity of a projectile changes by 9.8 m/s each second,
  • The horizontal motion of a projectile is independent of its vertical motion.
In this portion you will learn how to describe the motion of projectiles numerically. You will learn how the numerical values of the x- and y-components of the velocity and displacement change with time (or remain constant). As you proceed through this part , pay careful attention to how a conceptual understanding of projectiles translates into a numerical understanding.
Consider again the cannonball launched by a cannon from the top of a very high cliff. Suppose that the cannonball is launched horizontally with no upward angle whatsoever and with an initial speed of 20 m/s. If there were no gravity, the cannonball would continue in motion at 20 m/s in the horizontal direction. Yet in actuality, gravity causes the cannonball to accelerate downwards at a rate of 9.8 m/s/s. This means that the vertical velocity is changing by 9.8 m/s every second. If a vector diagram (showing the velocity of the cannonball at 1-second intervals of time) is used to represent how the x- and y-components of the velocity of the cannonball is changing with time, then x- and y- velocity vectors could be drawn and their magnitudes labeled. The lengths of the vector arrows are representative of the magnitudes of that quantity. Such a diagram is shown below.

 

The important concept depicted in the above vector diagram is that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second. These same two concepts could be depicted by a table illustrating how the x- and y-component of the velocity vary with time.

Time

Horizontal

Velocity

Vertical

Velocity

0 s
20 m/s, right
0
1 s
20 m/s, right
9.8 m/s, down
2 s
20 m/s, right
19.6 m/s, down
3 s
20 m/s, right
29.4 m/s, down
4 s
20 m/s, right
39.2 m/s, down
5 s
20 m/s, right
49.0 m/s, down

 

The numerical information in both the diagram and the table above illustrate identical points - a projectile has a vertical acceleration of 9.8 m/s/s, downward and no horizontal acceleration. This is to say that the vertical velocity changes by 9.8 m/s each second and the horizontal velocity never changes. This is indeed consistent with the fact that there is a vertical force acting upon a projectile but no horizontal force. A vertical force causes a vertical acceleration - in this case, an acceleration of 9.8 m/s/s.
But what if the projectile is launched upward at an angle to the horizontal? How would the horizontal and vertical velocity values change with time? How would the numerical values differ from the previously shown diagram for a horizontally launched projectile? The diagram below reveals the answers to these questions. The diagram depicts an object launched upward with a velocity of 75.7 m/s at an angle of 15 degrees above the horizontal. For such an initial velocity, the object would initially be moving 19.6 m/s, upward and 73.1 m/s, rightward. These values are x- and y-components of the initial velocity and will be discussed in more detail in the next part of this lesson.

Again, the important concept depicted in the above diagram is that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second. These same two concepts could be depicted by a table illustrating how the x- and y-component of the velocity vary with time.

Time

Horizontal

Velocity

Vertical

Velocity

0 s
73.1 m/s, right
19.6 m/s, up
1 s
73.1 m/s, right
9.8 m/s, up
2 s
73.1 m/s, right
0 m/s
3 s
73.1 m/s, right
9.8 m/s, down
4 s
73.1 m/s, right
19.6 m/s, down
5 s
73.1 m/s, right
29.4 m/s, down
6 s
73.1 m/s, right
39.2 m/s, down
7 s
73.1 m/s, right
49.0 m/s, down

The numerical information in both the diagram and the table above further illustrate the two key principles of projectile motion - there is a horizontal velocity that is constant and a vertical velocity that changes by 9.8 m/s each second. As the projectile rises towards its peak, it is slowing down (19.6 m/s to 9.8 m/s to 0 m/s); and as it falls from its peak, it is speeding up (0 m/s to 9.8 m/s to 19.6 m/s to ...). Finally, the symmetrical nature of the projectile's motion can be seen in the diagram above: the vertical speed one second before reaching its peak is the same as the vertical speed one second after falling from its peak. The vertical speed two seconds before reaching its peak is the same as the vertical speed two seconds after falling from its peak. For non-horizontally launched projectiles, the direction of the velocity vector is sometimes considered + on the way up and - on the way down; yet the magnitude of the vertical velocity (i.e., vertical speed) is the same an equal interval of time on either side of its peak. At the peak itself, the vertical velocity is 0 m/s; the velocity vector is entirely horizontal at this point in the trajectory. These concepts are further illustrated by the diagram below for a non-horizontally launched projectile that lands at the same height as which it is launched.

The above diagrams, tables, and discussion pertain to how the horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory. Another vector quantity that can be discussed is the displacement. .


Describing Projectiles With Numbers:
(Horizontal and Vertical Displacement)

The previous diagrams, tables, and discussion pertain to how the horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory. Now we will investigate the manner in which the horizontal and vertical components of a projectile's displacement vary with time. As has already been discussed, the vertical displacement (denoted by the symbol y in the discussion below) of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity. Thus, the vertical displacement (y) of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion. This equation was discussed in Unit 1 of The Physics Classroom. The equation can be written as follows.

y = 0.5 • g • t2

(equation for vertical displacement for a horizontally launched projectile)
where g is -9.8 m/s/s and t is the time in seconds. The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance that a projectile falls if dropped from rest. It was also discussed earlier, that the force of gravity does not influence the horizontal motion of a projectile. The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally (vix) and the amount of time (t) that it has been moving horizontally. Thus, if the horizontal displacement (x) of a projectile were represented by an equation, then that equation would be written as

x = vix • t

The diagram below shows the trajectory of a projectile (in red), the path of a projectile released from rest with no horizontal velocity (in blue) and the path of the same object when gravity is turned off (in green). The position of the object at 1-second intervals is shown. In this example, the initial horizontal velocity is 20 m/s and there is no initial vertical velocity (i.e., a case of a horizontally launched projectile).
As can be seen in the diagram above, the vertical distance fallen from rest during each consecutive second is increasing (i.e., there is a vertical acceleration). It can also be seen that the vertical displacement follows the equation above (y = 0.5 • g • t2). Furthermore, since there is no horizontal acceleration, the horizontal distance traveled by the projectile each second is a constant value - the projectile travels a horizontal distance of 20 meters each second. This is consistent with the initial horizontal velocity of 20 m/s. Thus, the horizontal displacement is 20 m at 1 second, 40 meters at 2 seconds, 60 meters at 3 seconds, etc. This information is summarized in the table below.

Time

Horizontal

Displacement

Vertical

Displacement

0 s
0 m
0 m
1 s
20 m
-4.9 m
2 s
40 m
-19.6 m
3 s
60 m
-44.1 m
4 s
80m
-78.4 m
5 s
100 m
-122.5 m

Now consider displacement values for a projectile launched at an angle to the horizontal (i.e., a non-horizontally launched projectile). How will the presence of an initial vertical component of velocity affect the values for the displacement? The diagram below depicts the position of a projectile launched at an angle to the horizontal. The projectile still falls 4.9 m, 19.6 m, 44.1 m, and 78.4 m below the straight-line, gravity-free path. These distances are indicated on the diagram below.

The projectile still falls below its gravity-free path by a vertical distance of 0.5*g*t^2. However, the gravity-free path is no longer a horizontal line since the projectile is not launched horizontally. In the absence of gravity, a projectile would rise a vertical distance equivalent to the time multiplied by the vertical component of the initial velocity (viy• t). In the presence of gravity, it will fall a distance of 0.5 • g • t2. Combining these two influences upon the vertical displacement yields the following equation.

y = viy • t + 0.5 • g • t2

(equation for vertical displacement for an angled-launched projectile)
where viy is the initial vertical velocity in m/s, t is the time in seconds, and g = -9.8 m/s/s (an approximate value of the acceleration of gravity). If a projectile is launched with an initial vertical velocity of 19.6 m/s and an initial horizontal velocity of 33.9 m/s, then the x- and y- displacements of the projectile can be calculated using the equations above. A sample calculation is shown below.

Calculations for t = 1 second

y = viy * t + 0.5*g*t2 where viy = 19.6 m/s
y = (19.6 m/s) * (1 s) + 0.5*(-9.8 m/s/s)*(1 s)2
y = 19.6 m + (-4.9 m)
y = 14.7 m (approximately)
x = vix * t where vix = 33.9 m/s
x = (33.9 m/s) * (1 s)
x = 33.9 m

The following table lists the results of such calculations for the first four seconds of the projectile's motion.

Time

Horizontal

Displacement

Vertical

Displacement

0 s
0 m
0 m
1 s
33.9 m
14.7 m
2 s
67.8 m
19.6 m
3 s
101.7 m
14.7 m
4 s
135.6 m
0 m
The data in the table above show the symmetrical nature of a projectile's trajectory. The vertical displacement of a projectile t seconds before reaching the peak is the same as the vertical displacement of a projectile t seconds after reaching the peak. For example, the projectile reaches its peak at a time of 2 seconds; the vertical displacement is the same at 1 second (1 s before reaching the peak) is the same as it is at 3 seconds (1 s after reaching its peak). Furthermore, the time to reach the peak (2 seconds) is the same as the time to fall from its peak (2 seconds).


 

Check Your Understanding

Use your understanding of projectiles to answer the following questions. Then click the button to view the answers.
1. Anna Litical drops a ball from rest from the top of 78.4-meter high cliff. How much time will it take for the ball to reach the ground and at what height will the ball be after each second of motion?
 
Click here to see a diagram of the situation.

2. A cannonball is launched horizontally from the top of an 78.4-meter high cliff. How much time will it take for the ball to reach the ground and at what height will the ball be after each second of travel?
 
Click here to see a diagram of the situation.

3. Fill in the table below indicating the value of the horizontal and vertical components of velocity and acceleration for a projectile.




4. The diagram below shows the trajectory for a projectile launched non-horizontally from an elevated position on top of a cliff. The initial horizontal and vertical components of the velocity are 8 m/s and 19.6 m/s respectively. Positions of the object at 1-second intervals are shown. Determine the horizontal and vertical velocities at each instant shown in the diagram.




The following diagram pertains to questions #1 and #2 above. A scale is used where 1 cm = 5 meters. (Note that 1-cm may be a different distance for different computer monitors; thus, a cm-ruler is given in the diagram.)
Return to Question #1.
Return to Question #2.



Initial Velocity Components

It has already been stated and thoroughly discussed that the horizontal and vertical motions of a projectile are independent of each other. The horizontal velocity of a projectile does not affect how far (or how fast) a projectile falls vertically. Perpendicular components of motion are independent of each other. Thus, an analysis of the motion of a projectile demands that the two components of motion are analyzed independent of each other, being careful not to mix horizontal motion information with vertical motion information. That is, if analyzing the motion to determine the vertical displacement, one would use kinematic equations with vertical motion parameters (initial vertical velocity, final vertical velocity, vertical acceleration) and not horizontal motion parameters (initial horizontal velocity, final horizontal velocity, horizontal acceleration). It is for this reason that one of the initial steps of a projectile motion problem is to determine the components of the initial velocity.

Earlier in this unit, the method of vector resolution was discussed. Vector resolution is the method of taking a single vector at an angle and separating it into two perpendicular parts. The two parts of a vector are known as components and describe the influence of that vector in a single direction. If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component. The horizontal velocity component (vx) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component (vy) describes the influence of the velocity in displacing the projectile vertically. Thus, the analysis of projectile motion problems begins by using the trigonometric methods discussed earlier to determine the horizontal and vertical components of the initial velocity.

Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal. Such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity. These numerical values were determined by constructing a sketch of the velocity vector with the given direction and then using trigonometric functions to determine the sides of the velocity triangle. The sketch is shown at the right and the use of trigonometric functions to determine the magnitudes is shown below. (If necessary, review this method on an earlier page in this unit.)

 


All vector resolution problems can be solved in a similar manner. As a test of your understanding, utilize trigonometric functions to determine the horizontal and vertical components of the following initial velocity values. When finished, click the button to check your answers.

Practice A: A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.
 

Practice B: A motorcycle stunt person traveling 70 mi/hr jumps off a ramp at an angle of 35 degrees to the horizontal.
 
Practice C: A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.
 


Try Some More!

Need more practice? Use the Velocity Components for a Projectile widget below to try some additional problems. Enter any velocity magnitude and angle with the horizontal. Use your calculator to determine the values of vx and vy. Then click the Submit button to check your answers.

Velocity Components for a Projectile

Velocity magnitude:
m/s
Angle w/horizontal:
deg



Submit






As mentioned above, the point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters as the horizontal displacement, the vertical displacement, the final vertical velocity, the time to reach the peak of the trajectory, the time to fall to the ground, etc. This process is demonstrated on the remainder of this page. We will begin with the determination of the time.

 

Determination of the Time of Flight

The time for a projectile to rise vertically to its peak (as well as the time to fall from the peak) is dependent upon vertical motion parameters. The process of rising vertically to the peak of a trajectory is a vertical motion and is thus dependent upon the initial vertical velocity and the vertical acceleration (g = 9.8 m/s/s, down). The process of determining the time to rise to the peak is an easy process - provided that you have a solid grasp of the concept of acceleration. When first introduced, it was said that acceleration is the rate at which the velocity of an object changes. An acceleration value indicates the amount of velocity change in a given interval of time. To say that a projectile has a vertical acceleration of -9.8 m/s/s is to say that the vertical velocity changes by 9.8 m/s (in the - or downward direction) each second. For example, if a projectile is moving upwards with a velocity of 39.2 m/s at 0 seconds, then its velocity will be 29.4 m/s after 1 second, 19.6 m/s after 2 seconds, 9.8 m/s after 3 seconds, and 0 m/s after 4 seconds. For such a projectile with an initial vertical velocity of 39.2 m/s, it would take 4 seconds for it to reach the peak where its vertical velocity is 0 m/s. With this notion in mind, it is evident that the time for a projectile to rise to its peak is a matter of dividing the vertical component of the initial velocity (viy) by the acceleration of gravity.

Once the time to rise to the peak of the trajectory is known, the total time of flight can be determined. For a projectile that lands at the same height which it started, the total time of flight is twice the time to rise to the peak. Recall from the last section of Lesson 2 that the trajectory of a projectile is symmetrical about the peak. That is, if it takes 4 seconds to rise to the peak, then it will take 4 seconds to fall from the peak; the total time of flight is 8 seconds. The time of flight of a projectile is twice the time to rise to the peak.

Determination of Horizontal Displacement

The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. As discussed in the previous part of this lesson, the horizontal displacement of a projectile can be determined using the equation

x = vix • t

If a projectile has a time of flight of 8 seconds and a horizontal velocity of 20 m/s, then the horizontal displacement is 160 meters (20 m/s • 8 s). If a projectile has a time of flight of 8 seconds and a horizontal velocity of 34 m/s, then the projectile has a horizontal displacement of 272 meters (34 m/s • 8 s). The horizontal displacement is dependent upon the only horizontal parameter that exists for projectiles - the horizontal velocity (vix).

Determination of the Peak Height

A non-horizontally launched projectile with an initial vertical velocity of 39.2 m/s will reach its peak in 4 seconds. The process of rising to the peak is a vertical motion and is again dependent upon vertical motion parameters (the initial vertical velocity and the vertical acceleration). The height of the projectile at this peak position can be determined using the equation

y = viy • t + 0.5 • g • t2

where viy is the initial vertical velocity in m/s, g is the acceleration of gravity (-9.8 m/s/s) and t is the time in seconds it takes to reach the peak. This equation can be successfully used to determine the vertical displacement of the projectile through the first half of its trajectory (i.e., peak height) provided that the algebra is properly performed and the proper values are substituted for the given variables. Special attention should be given to the facts that the t in the equation is the time up to the peak and the g has a negative value of -9.8 m/s/s.


Check Your Understanding


Answer the following questions and click the button to see the answers.
1. Aaron Agin is resolving velocity vectors into horizontal and vertical components. For each case, evaluate whether Aaron's diagrams are correct or incorrect. If incorrect, explain the problem or make the correction.



2. Use trigonometric functions to resolve the following velocity vectors into horizontal and vertical components. Then utilize kinematic equations to calculate the other motion parameters. Be careful with the equations; be guided by the principle that "perpendicular components of motion are independent of each other."
 
 
 


3. Utilize kinematic equations and projectile motion concepts to fill in the blanks in the following tables.



Horizontally Launched Projectile Problems

One of the powers of physics is its ability to use physics principles to make predictions about the final outcome of a moving object. Such predictions are made through the application of physical principles and mathematical formulas to a given set of initial conditions. In the case of projectiles, a student of physics can use information about the initial velocity and position of a projectile to predict such things as how much time the projectile is in the air and how far the projectile will go. The physical principles that must be applied are those discussed previously in Lesson 2. The mathematical formulas that are used are commonly referred to as kinematic equations. Combining the two allows one to make predictions concerning the motion of a projectile. In a typical physics class, the predictive ability of the principles and formulas are most often demonstrated in word story problems known as projectile problems.
There are two basic types of projectile problems that we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are:

Problem Type 1:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.
Examples of this type of problem are
  1. A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
  2. A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Problem Type 2:
A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.
Examples of this type of problem are
  1. A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
  2. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2. In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:


Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are
Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax = 0 m/s/s.

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are
In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), viy = 0 m/s. Thus, any term with viy in it will cancel out of the equation.
The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

Example

A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

Horizontal Information

Vertical Information

x = ??? vix = 2.4 m/s
ax = 0 m/s/s
y = -0.60 m viy = 0 m/s
ay = -9.8 m/s/s

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = viy•t +0.5•ay•t2) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

-0.60 m = (0 m/s)•t + 0.5•(-9.8 m/s/s)•t2

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

-0.60 m = (-4.9 m/s/s)•t2

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

0.122 s2 = t2

By taking the square root of both sides of the equation, the time of flight can then be determined.

t = 0.350 s (rounded from 0.3499 s)

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information, vix = 2.4 m/s and ax = 0 m/s/s. The first horizontal equation (x = vix•t + 0.5•ax•t2) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

x = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

x = (2.4 m/s)•(0.3499 s)

Thus,

x = 0.84 m (rounded from 0.8398 m)

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

 

The following procedure summarizes the above problem-solving approach.
  1. Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
  2. Identify the unknown quantity that the problem requests you to solve for.
  3. Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
  4. With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)
One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold. Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

 


Check Your Understanding

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
 


Non-Horizontally Launched Projectile Problems

In the previous part of Lesson 2, the use of kinematic equations to solve projectile problems was introduced and demonstrated. These equations were used to solve problems involving the launching of projectiles in a horizontal direction from an elevated position. In this section of Lesson 2, the use of kinematic equations to solve non-horizontally launched projectiles will be demonstrated. A non-horizontally launched projectile is a projectile that begins its motion with an initial velocity that is both horizontal and vertical. To treat such problems, the same principles that were discussed earlier in Lesson 2 will have to be combined with the kinematic equations for projectile motion. You may recall from earlier that there are two sets of kinematic equations - a set of equations for the horizontal components of motion and a similar set for the vertical components of motion. For the horizontal components of motion, the equations are
Of these three equations, the top equation is the most commonly used. The other two equations are seldom (if ever) used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax = 0 m/s/s.
For the vertical components of motion, the three equations are
In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity).
As discussed earlier in Lesson 2, the vix and viy values in each of the above sets of kinematic equations can be determined by the use of trigonometric functions. The initial x-velocity (vix) can be found using the equation vix = vi•cosine(Theta) where Theta is the angle that the velocity vector makes with the horizontal. The initial y-velocity (viy) can be found using the equation viy = vi•sine(Theta) where Theta is the angle that the velocity vector makes with the horizontal. The topic of components of the velocity vector was discussed earlier in Lesson 2.

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider their use in the solution of the following problem.

Example A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.
The solution of any non-horizontally launched projectile problem (in which vi and Theta are given) should begin by first resolving the initial velocity into horizontal and vertical components using the trigonometric functions discussed above. Thus,

Horizontal Component

Vertical Component

vix = vi•cos(Theta) vix = 25 m/s•cos(45 deg)
vix = 17.7 m/s
viy = vi•sin(Theta) viy = 25 m/s•sin(45 deg)
viy = 17.7 m/s
In this case, it happens that the vix and the viy values are the same as will always be the case when the angle is 45-degrees.
The solution continues by declaring the values of the known information in terms of the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. In this case, the following information is either explicitly given or implied in the problem statement:

Horizontal Information

Vertical Information

x = ??? vix = 17.7 m/s
vfx = 17.7 m/s
ax = 0 m/s/s
y = ??? viy = 17.7 m/s
vfy = -17.7 m/s
ay = -9.8 m/s/s

As indicated in the table, the final x-velocity (vfx) is the same as the initial x-velocity (vix). This is due to the fact that the horizontal velocity of a projectile is constant; there is no horizontal acceleration. The table also indicates that the final y-velocity (vfy) has the same magnitude and the opposite direction as the initial y-velocity (viy). This is due to the symmetrical nature of a projectile's trajectory.
The unknown quantities are the horizontal displacement, the time of flight, and the height of the football at its peak. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. There are a variety of possible strategies for solving the problem. An organized listing of known quantities in two columns of a table provides clues for the selection of a useful strategy.
From the vertical information in the table above and the second equation listed among the vertical kinematic equations (vfy = viy + ay*t), it becomes obvious that the time of flight of the projectile can be determined. By substitution of known values, the equation takes the form of
-17.7 m/s = 17.7 m/s + (-9.8 m/s/s)•t
The physics problem now takes the form of an algebra problem. By subtracting 17.7 m/s from each side of the equation, the equation becomes
-35.4 m/s = (-9.8 m/s/s)•t
If both sides of the equation are divided by -9.8 m/s/s, the equation becomes
3.61 s = t (rounded from 3.6077 s)
The total time of flight of the football is 3.61 seconds.
With the time determined, information in the table and the horizontal kinematic equations can be used to determine the horizontal displacement (x) of the projectile. The first equation (x = vix•t + 0.5•ax•t2) listed among the horizontal kinematic equations is suitable for determining x. With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of
x = (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s)2
Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to
x = (17.7 m/s)•(3.6077 s)
Thus,
x = 63.8 m
The horizontal displacement of the projectile is 63.8 m.
Finally, the problem statement asks for the height of the projectile at is peak. This is the same as asking, "what is the vertical displacement (y) of the projectile when it is halfway through its trajectory?" In other words, find y when t = 1.80 seconds (one-half of the total time). To determine the peak height of the projectile (y with t = 1.80 sec), the first equation (y = viy•t +0.5•ay•t2) listed among the vertical kinematic equations can be used. By substitution of known values into this equation, it takes the form of
y = (17.7 m/s)•(1.80 s) + 0.5*(-10 m/s/s)•(1.80 s)2
Using a calculator, this equation can be simplified to
y = 31.9 m + (-15.9 m)
And thus,
y = 15.9 m
The solution to the problem statement yields the following answers: the time of flight of the football is 3.61 s, the horizontal displacement of the football is 63.8 m, and the peak height of the football 15.9 m.
(Note that in all calculations performed above, unrounded numbers were used. The numbers reported in the preliminary steps and in the final answer were the rounded form of the actual unrounded values.)

The following procedure summarizes the above problem-solving approach.
  1. Use the given values of the initial velocity (the magnitude and the angle) to determine the horizontal and vertical components of the velocity (vix and viy).
  2. Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
  3. Identify the unknown quantity that the problem requests you to solve for.
  4. Select either a horizontal or vertical equation to solve for the time of flight of the projectile. For non-horizontally launched projectiles, the second equation listed among the vertical equations (vfy = viy + ay*t) is usually the most useful equation.
  5. With the time determined, use a horizontal equation (usually x = vix*t + 0.5*ax*t2 ) to determine the horizontal displacement of the projectile.
  6. Finally, the peak height of the projectile can be found using a time value that is one-half the total time of flight. The most useful equation for this is usually y = viy*t +0.5*ay*t2 .
One caution is in order: the sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! And problems can often be simplified by the use of short procedures as the one above. However, not all problems can be solved with the above procedure. While steps 1, 2 and 3 above are critical to your success in solving non-horizontally launched projectile problems, there will always be a problem that doesn't "fit the mold." Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.



Your Turn to Try It!

Use the Range of an Angle-Launched Projectile widget to practice a projectile problem (or two) (or three). Using the given launch velocity and launch angle, determine the expected horizontal displacement (dx). After completing your calculation, use the Submit button to check your answer.


Range of an Angle-Launched Projectile

Launch Velocity:
m/s
Launch Angle:
°



Submit
See http://www.physicsclassroom.com/Class/vectors/U3L2a.cfm.






Check Your Understanding

A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.



adapted from : http://www.physicsclassroom.com
 

Copyright © 2010 A Level Information Hub | Premium Blogger Templates & Photography Logos | PSD Design by Amuki