Projectile Motion
Chapter 3 Section 3
What is Projectile Motion?

Projectile Motion – Motion that is launched into
the air that is subject to gravity and described in
two dimensions.

Examples of projectiles:
–
–
–
–
–
baseballs
footballs
bullets
arrows
etc….
How to describe Projectiles

Projectile Motion is motion in 2dimensions.

When solving for problems dealing with 2dimensional motion, it is best to break the
motion into 1-dimensional parts
– (Vertical and Horizontal)

Once solved in 1-dimensional, recombine
the components to find the final resultant.
Kinematic Equations
The kinematic equations are still used to
solve for projectile motion and are applied
in one dimension at a time.
 The setup is the same, but different
variables are used to help expression the
motion in the x- and y-directions…

Kinematic Variable
Here is the setup for the variables:
𝑣𝑖𝑥 =
𝑣𝑖𝑦 =

𝑡=
𝑣𝑓𝑦 =
𝑑𝑥 =
g=
𝑡=
𝑑𝑦 =
Trajectories

Objects that are in a projectile motion
follow parabolic trajectories
– Figure 3-18 pg 99 in book shows a great
example…
Horizontal Motion of Projectiles
Objects that have an initial horizontal velocity
retain that velocity as the objects continues in its
parabolic trajectory.
 Example:

– If a person runs off a cliff with a velocity of 20m/s,
that person will continue to move at 20m/s
horizontally as the person falls to the ground below.

Horizontal velocity is considered a constant in
projectile problems.
Vertical Motion of Projectiles
As an object is in projectile motion, it
continues to have gravity acting on it and
falls towards the earth at an acceleration
of 9.8m/s² straight downward.
 Projectile motion is nothing more than
free fall with an initial horizontal velocity.

– Figure 3-19 pg 99 in book
Components of Projectiles

Breaking projectile motion in to
components can greatly simplify the
problem.
– Motion can be described in the x-direction and
the y-direction.
Vix
Viy
Vi
Vector Diagram
Viy
Vi
θ
Vix
Finding the Components

The sine and cosine functions can be used to find
the horizontal and vertical components of the
initial velocity.
𝑣𝑖𝑥 = 𝑣𝑖 𝑐𝑜𝑠𝜃
𝑣𝑖𝑦 = 𝑣𝑖 𝑠𝑖𝑛𝜃
Projectile Cases

There are 3 different cases in which a
projectile can be described.
– Case 1: Object with only horizontal velocity and
no vertical velocity falling with negative vertical
displacement.
– Case 2: Object that is shot upward at some
angle and has both horizontal and vertical
velocity and lands with zero vertical
displacement.
– Case 3: Object that is shot at some angle and
has both horizontal and vertical velocity with
negative or positive vertical displacement.
Kinematic Equations for Case 1
Projectiles
Vertical motion of a
projectile
1 2
𝑔𝑡
2
𝑑𝑦 =
𝑣𝑓𝑦 = 𝑔𝑡
2
𝑣𝑓𝑦 = 2𝑔𝑑𝑦
Horizontal motion of
a projectile
𝑑𝑥 = 𝑣𝑖𝑥 𝑡
𝑣𝑓𝑥 = 𝑣𝑖𝑥
Example Problem #1

A car is traveling at 37.0 km/hr on a
perfectly horizontal road when it suddenly
loses control and runs off a cliff which is
17.30 meters tall. How far did the car
travel before crashing into the ground
below the cliff?
Example Problem #1 Answer

dx = 19.33m
Projectiles Launched at an Angle

Projectiles are mostly launched at some
angle to the horizontal in real-world
application.
– Examples
 Bullets
 Footballs
 Baseballs

The projectile has an initial vertical
component of velocity as well as a
horizontal component of velocity.
Maximum Range

To achieve maximum range of a projectile,
it should be fired at a 45 degree angle to
the horizontal.
Case 2 Equations

With some algebra and trigonometry, the
kinematic equations can be rearranged to
solve for certain situations.
Special Case 2 Equations
𝑦𝑚𝑎𝑥
𝑣𝑖 𝑠𝑖𝑛𝜃
=−
2𝑔
𝑡𝑟𝑎𝑛𝑔𝑒
2
2𝑣𝑖 𝑠𝑖𝑛𝜃
=−
𝑔
𝑡max ℎ𝑒𝑖𝑔ℎ𝑡
𝑣𝑖 𝑠𝑖𝑛𝜃
=−
𝑔
2
𝑣𝑖 𝑠𝑖𝑛2𝜃
𝑅𝑎𝑛𝑔𝑒 = −
𝑔
Example Problem #2

A quarterback throws a football with a
velocity of 27.50m/s at an angle of 35
degrees above the horizontal.
a)
b)
c)
d)
What is the maximum height?
What is the maximum range?
How long is the football in the air?
What is the impact speed of the football
hitting the ground?
Example Problem #2 Answer
a)
b)
c)
d)
12.69 m
72.52 m
3.22 s
52.42 m/s
Kinematic Equations for Projectile
Motion
Vertical motion of a
projectile
1 2
𝑔𝑡
2
𝑑𝑦 = 𝑣𝑖𝑦 𝑡 +
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
𝑣𝑓𝑦 2 = 𝑣𝑖𝑦 2 + 2𝑔𝑑𝑦
Horizontal motion of a
projectile
𝑑𝑥 = 𝑣𝑖𝑥 𝑡
𝑣𝑓𝑥 = 𝑣𝑖𝑥
Impact Velocity and Speed

The velocity, or speed, as an object strikes
the ground is a combination of Vfx and Vfy.
𝐼𝑚𝑝𝑎𝑐𝑡 𝑆𝑝𝑒𝑒𝑑 =
𝑣𝑓𝑥 2 + 𝑣𝑓𝑦 2
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = tan
−1
𝑣𝑓𝑦
𝑣𝑓𝑥
Example Problem #3

A person throws a ball with a velocity of
23.40 m/s at 55 degrees above the
horizontal to a friend on top of a small
building, which is 21.70 m tall. If the
person is standing 24.0 meters away from
the building on the ground, will the ball
make it over the top of the building and
onto the roof?
Example Problem #3 Answer

No, the ball does not make it to the top of
the roof. The ball only goes 18.61m high
and the building is 21.70m tall.
Example Problem #4

In a scene in a action movie, a stuntman
jumps from the top of one building to the
top of another building 4.0m away. After
a running start, he leaps at an angle of
15º with respect to the flat floor while
traveling at a speed of 5.0m/s. Will he
make it to the other roof, which is 2.5m
shorter than the building he jumps from?
Example Problem #4 Answer
4.13 m jump across the buildings
 Yes, he makes the jump.
