Projectile Motion

What is projectile?
◼ Projectile is any object which projected by some means
and continues to move due to its own inertia (mass).
◼ The path of a projectile is called its trajectory.
◼ The force of primary importance acting on a projectile is
gravity
◼ A projectile is any object with an initial non-zero.
Horizontal velocity whose acceleration is due to gravity
alone
◼ The kinematic equations for a simple projectile are those
of an object traveling at constant horizontal velocity and
constant vertical acceleration.
◼ The trajectory of a simple projectile is a parabola
What is projectile?
◼ Projectilemotion can be
described as the motion of an
object (called the projectile) that is
moving in the x-y plane. This just
means that the motion of the
projectile is the vector sum of its
horizontal and vertical motion.
Examples of projectile?

◼ A baseball that has been pitched, batted, or thrown

◼ A bullet the instant in exits the barrel of a gun or rifle
◼ A bus drove off an uncompleted bridge
◼ A moving airplane in the air with its engines and
wings disabled
◼ A runner in mid-stride (since they momentarily lose
contact with the ground)
◼ The space shuttle or any other spacecraft after the
main engine cut-ff
Projectiles move in TWO dimensions
Since a projectile
moves in 2-
dimensions, it
therefore has 2
components just
like a resultant
vector.
◼ Horizontal and
Vertical
Horizontal “Velocity” Component

◼ NEVER changes, covers equal displacements in

equal time periods. This means the initial
horizontal velocity equals the final horizontal
velocity

In other words, the horizontal

velocity is CONSTANT. BUT
WHY?

Gravity DOES NOT work

horizontally to increase or
decrease the velocity.
Vertical “Velocity” Component
◼ Changes (due to gravity), does NOT cover
equal displacements in equal time periods.

Both the MAGNITUDE and DIRECTION change. As

the projectile moves up the MAGNITUDE
DECREASES and its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and the
direction is DOWNWARD.
Combining the Components
Together, these
components produce
what is called a
trajectory or path. This
path is parabolic in
nature.

Component Magnitude Direction

Horizontal Constant Constant
Vertical Changes Changes
Combining the Components
◼ Along the horizontal, the velocity is
constant and free from acceleration (a =
0). The acceleration is only present in the
vertical. The kinematic equations still apply
in projectile motion. A point to consider in
projectile motion is that at its maximum
height, the vertical component of the
velocity becomes zero.
Kinematic Equation for Projectile Motion

Horizontal Motion Vertical Motion

ax = 0 vx – constant ay = -ag = constant

vfx = vix vfy = viy - agt

xf = xi + vixt yf = yi + viyt – ½ ag

vfy2 = viy2 – 2ag (yf – yi)

 Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial
VERTICAL velocity.
vox = vx = constant

voy = 0 m / s
Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for
the “y” direction. And for this we use kinematic #2.

x = vox t + 1 at 2
2
x = vox t y = 1 gt 2
2
Remember, the velocity is Remember that since the
CONSTANT horizontally, so projectile is launched
that means the acceleration horizontally, the INITIAL
is ZERO! VERTICAL VELOCITY is
equal to ZERO.
Horizontally Launched Projectiles
Example: A plane traveling with What do I What I want to
a horizontal velocity of 100 know? know?
m/s is 500 m above the
ground. At some point the vox=100 m/s t=?
pilot drops a bomb on a
target below. (a) How long is y = 500 m x=?
the bomb in the air? (b) How
far away from point above voy= 0 m/s
where it was dropped will it
land? g = 9.8 m/s/s
Horizontally Launched Projectiles

y = 1 gt 2 → −500 = 1 (−9.8)t 2
2 2
x = vox t = (100)(10.1) = 1010 m
102.04 = t 2 → t = 10.1 seconds
Seatwork
◼ A cannon ball is fired with a horizontal
velocity of 300 m/s from the top of a cliff 60 m
high. How far from the base of cliff will it hit
the ground?
Example 1
◼ A baseball is thrown upward with a speed of 20 m/s
at an angle of 40o from the horizontal. Determine
the horizontal and vertical components of its
velocity.
Solution:
◼ From our review of trigonometry and vector
resolution, we know that the horizontal component
of the vector is given by the cosine function and the
vertical component by the sine function. Hence,
vx =vcosθ
vy =vsinθ
Example 1
◼ From the problem, the following data were
gleaned:
v=20ms θ = 40°
◼ Therefore, the horizontal and vertical
components of the baseball’s velocity are
vx =vcosθ =(20m/s)cos(40°)
vx =15.321m/s
vy =vsinθ =(20ms)sin(40°)
vy =12.856m/s
Vertically Launched Projectiles
NO Vertical Velocity at the top of the trajectory.

Vertical Vertical Velocity

Velocity increases on the
decreases way down,
on the way
upward Horizontal Velocity
is constant

Component Magnitude Direction

Horizontal Constant Constant
Vertical Decreases up, 0 Changes
@ top, Increases
down
Vertically Launched Projectiles
Since the projectile was launched at an angle, the
velocity MUST be broken into components!!!

vox = vo cos q
vo voy
voy = vo sin q
q
vox
Vertically Launched Projectiles
There are several
things you must
consider when doing
these types of
projectiles besides
using components. If
it begins and ends at
ground level, the “y”
displacement is
ZERO: y = 0
Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.

vo voy x = voxt y = voy t + 1 gt 2

2
q
vox vox = vo cos q
voy = vo sin q
Example
A place kicker kicks a football with a velocity of 20.0 m/s
and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?

vox = vo cos q
vox = 20 cos 53 = 12.04 m / s
voy = vo sin q
q = 53
voy = 20sin 53 = 15.97 m / s
Example
A place kicker kicks a What I know What I want
football with a to know
velocity of 20.0 m/s vox=12.04 m/s t=?
and at an angle of 53 voy=15.97 m/s x=?
degrees.
y=0 ymax=?
(a) How long is the ball
in the air? g = - 9.8
m/s/s

y = voy t + 1 gt 2 → 0 = (15.97)t − 4.9t 2

2
−15.97t = −4.9t → 15.97 = 4.9t
2

t = 3.26 s
Example

A place kicker kicks a What I know What I want

football with a to know
velocity of 20.0 m/s vox=12.04 m/s t = 3.26 s
and at an angle of 53 voy=15.97 m/s x=?
degrees. y=0 ymax=?
(b) How far away does it g = - 9.8
land? m/s/s

x = vox t → (12.04)(3.26) = 39.24 m

Example
What I know What I want
to know
A place kicker kicks a vox=12.04 m/s t = 3.26 s
football with a velocity voy=15.97 m/s x = 39.24 m
of 20.0 m/s and at an
angle of 53 degrees. y=0 ymax=?
g = - 9.8
(c) How high does it m/s/s
travel?
y = voy t + 1 gt 2
2
CUT YOUR TIME IN HALF! y = (15.97)(1.63) − 4.9(1.63) 2

y = 13.01 m