
In this packet we will:
 Define projectile motion.
 Discuss how displacement, velocity, and
acceleration are related to this type of motion.
 Provide several examples for the student to
familiarize themselves with projectile motion.
In physics, remember that we always try to define everything as precisely as we
possibly can, so before studying projectile motion we first need to understand what we
mean by a projectile.
Definition:
Projectile: An object moving in space with only the force of gravity acting on it.
Note: Now when I say “space” here I do not mean outer-space and space ships. In
physics we often use the term space to represent something an object can move
through. That may be the air, water, outer-space, etc., but most often than not it’s air.
Q: Can you think of an object moving through space with only gravity acting on it?
Let’s take a look at some examples to clarify that definition.
Example 1:
Someone dropping a ball off a building. The path the ball takes, if we neglect air
resistance, can be considered projectile motion.
This problems is actually quite similar to Example
Problem 1 from the packet Acceleration Part 2.
The only difference is that this time we have a
ball and not a penny.
Another interesting note is that we can even use
the same governing equations developed in the
packets Acceleration Part 2 and Acceleration Part
3 to describe this and all the other projectile
motion problems we will be discussing. More on
this later.
Example 2:
A soccer player kicking a soccer ball through the air.
The path the soccer ball travels through (the dashed line) is also given a special name.
Definition:
Trajectory: The path a projectile follows.
At this point you should also notice that the trajectory resembles a parabola. This fact
might allow us to use concepts from algebra to help determine information about a
projectile. More on this later.
Now that we have seen some example of projectile motion, let’s now start to relate
our concepts of displacement, velocity, and acceleration to it.
From previous packets on acceleration we developed an equation which relates
displacement, velocity, and acceleration. That equation is the following:
1
2
at   vi t  ri  rf
2
The big difference between how we used this equation before and how we are
going to use it now it that before we used it to solve 1-dimensional problems. But as
you can see from Example 2, projectile motion is a two dimensional problem.
But fear not! Since all the quantities we are dealing with are vectors, we know from
our definition of vectors that we can break vectors into x and y components.
Therefore, if we did this here instead of dealing with a 2-dimensional problem we
would be dealing with 2 1-dmensional problems.
Then we can just proceed like we have done previously.
So the question then is, how do we turn the equation on the previous slide into two
separate equations?
If we recall from our knowledge of vectors, that the components of vectors are
independent of one another, we can say the following.

 
1
2
a t   vi t  ri  rf
2
1
2
a x t   vi , x t  ri , x  rf , x
2
1
2
a y t   vi , y t  ri , y  rf , y
2
This tells us that if we were to plot the motion of a projectile on an xy-coordinate
plane, the horizontal motion of the projectile (motion along the x-axis) would be
independent of the vertical motion of the projectile (motion along the y-axis).
You may try to prove this rule for yourself, by trying the following experiment.
Experiment:
Drop a ball from the edge of a table top while at the same instant rolling another ball
off the same table top. Which ball hits the floor first?
Now let’s move on to simplifying the equations on Slide 6.
Remember we said that a projectile is a object moving through space with only
acceleration due to gravity, g, acting on it. We also know that when we are close to
the Earth that we assume g is pointing downwards, when drawn on an xy-coordinate
plane.
We can then infer from this that we must have a vertical or y-component to the
acceleration. But, do we have an x-component for the acceleration. Or in other
words is there any acceleration in the x-direction?
The answer is no. There is now acceleration in the x-direction because only
acceleration due to gravity is acting on the object.
So what implications does this have for the equations on Slide 6? Well, let’s see.
Rewriting the equations we have:
1
2
a x t   vi , x t  ri , x  rf , x
2
1
2
a y t   vi , y t  ri , y  rf , y
2
We know that:
ax  0
ay  g
Substituting these values into their respective equations we find the following:
vi , x t  ri , x  rf , x
1
2
g t   vi , y t  ri , y  rf , y
2
Therefore, we now have equations which describe the horizontal position (x-position)
and the vertical position (y-position) of a projectile with time.
If we were to write the previous equations in terms of functions of time we get the
following:
x(t )  xi  vi , xt
1 2
y (t )  yi  vi , y t  gt
2
Note: For the y-position function the last term is negative because g is considered to
be pointing downwards in the xy-coordinate plane.
Note: In many textbooks vi, xi, and yi are written as vo, xo, and yo however since I
have been using vi, xi, and yi I will try to keep that convention throughout.
The first equation above describes the distance a projectile travels in the horizontal
with time, this is what we call a projectiles range.
Definition:
Range: The total horizontal distance a projectile travels.
Now that we have some equations which describe the motion of a projectile, let’s try
some example problems to make ourselves comfortable with using them.
At this point you might think to yourself, “So we know g, and we might be able to
measure positions, but how do we determine the components of the velocity for the x
and y directions?”
You should remember from the packets Vectors and Scalars Part 1 and 2 and some
knowledge of trigonometry, that the horizontal component (x-component) and the
vertical component (y-component) of velocity can be written as:
vi , x  vi cos 
vi , y  vi sin 
Our equations then become:
x(t )  xi  (vi cos  )t
1 2
y (t )  yi  (vi sin  )t  gt
2
Example 3:
During a soccer game a player kicks a ball through the air, with an initial velocity of 10
m/s at an angle of 30 degrees to the horizontal. How far will the soccer ball travel
down the soccer field? How long will the soccer ball take to land on the ground?
Example 4:
While playing catch you throw a football through the air at 30 m/s at an angle of 45
degrees. When does the football reach a height of 6 m? Assume you let the ball go at
a height of 2 m.
Example 5:
In a baseball game you step up to bat and get a hit. If the ball left your bat at a
velocity of 45 m/s at an angle of 45 degrees how far along the field will the ball travel?
If the homerun marker is at 100 meters and the wall surrounding the field is 2 meters
high, did you score a homerun? Assume you hit the ball at a height of 1 meter.