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
F
grav). 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
• t
2). 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 (v
iy• t).
In the presence of gravity, it will fall a distance of 0.5
• g • t
2. 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 v
x and v
y. Then click the
Submit button to check your answers.
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 (v
iy) 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
- 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.
- 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
- 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.
- 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 a
x 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),
v
iy = 0 m/s. Thus, any term with v
iy
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, v
ix, v
iy, a
x,
a
y, 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 = v
iy•t
+0.5•a
y•t
2) 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, v
ix = 2.4 m/s and a
x
= 0 m/s/s. The first horizontal equation (x =
v
ix•t +
0.5•a
x•t
2) 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.
- 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.
- Identify the unknown quantity that the problem
requests you to solve for.
- Select either a horizontal or vertical equation to
solve for the time of flight of the projectile.
- 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
a
x 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 v
ix and v
iy
values in each of the above sets of kinematic equations can
be determined by the use of trigonometric functions. The
initial x-velocity (v
ix) can be found using the
equation v
ix = v
i•cosine(Theta)
where Theta is the angle that the velocity vector makes
with the horizontal. The initial y-velocity (v
iy)
can be found using the equation v
iy =
v
i•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 v
i 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 v
ix and the
v
iy 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, v
ix, v
iy,
a
x, a
y, 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
(v
fx) is the same as the initial x-velocity
(v
ix). 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 (v
fy) has the same magnitude
and the opposite direction as the initial y-velocity
(v
iy). 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 (v
fy = v
iy +
a
y*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 = v
ix•t +
0.5•a
x•t
2) 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 = v
iy•t
+0.5•a
y•t
2) 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.
- 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).
- 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.
- Identify the unknown quantity that the problem
requests you to solve for.
- 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.
- With the time determined, use a horizontal equation
(usually x = vix*t + 0.5*ax*t2
) to determine the horizontal displacement of the
projectile.
- 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 (d
x). After completing your calculation, use the
Submit button to check your answer.
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
