Projectile Motion
A projectile is an
object moving in
two dimensions
under the influence
of Earth's gravity;
its path is an upside
down parabola.
Projectile Motion
• Projectile Motion  Motion of an object
that is projected into the air at an angle.
• Near the Earth’s surface, the acceleration a
on the projectile is downward and equal to
a = g = 9.8 m/s2
Goal: Describe projectile motion after it starts.
• Galileo: Analyzed horizontal & vertical
components of motion separately.
• Today: Displacement D & velocity v are vectors
 Components of motion can
be treated separately
Projectile Motion
• Simplest example: A ball rolls across a table, to
the edge & falls off the edge to the floor. It leaves
the table at time t = 0. Analyze the y part of
motion & the x part of motion separately.
• y part of the motion: Down is positive & the origin
is at table top: y0 = 0. Initially, there is no y component
of velocity: vy0 = 0
 vy = gt, y = (½)g t2
• x part of motion: The origin is at the table top: x0 = 0.
No x component of acceleration(!) ax = 0.
Initially the x component of velocity is vx0
 vx = vx0 , x = vx0t
A Ball Rolls Across Table & Falls Off
t = 0 here
Projectiles can be understood
by analyzing horizontal
vertical motions separately. At
any point, v has both x & y
components. Take down as
positive. Initial velocity has an
x component ONLY! That is
vy0 = 0. The kinematic
equations tell us that, at time t,
vx = vx0, vy = gt
x = vx0t
y = vy0t + (½)gt2
Summary
• A ball rolling across the table & falling.
• The vector velocity v has 2 components:
vx = vx0 , vy = gt
• Vector displacement D has 2 components:
x = vx0t , y = (½)g t2
• The speed in the x direction is
constant; in the y-direction the
object moves with constant
acceleration g.
• The photo shows 2 balls that
start to fall at the same time.
• The one on the right has an
initial speed in the x direction.
• It can be seen that the vertical
positions of the 2 balls are
identical at identical times,
while the horizontal position
of the yellow ball increases
linearly.
Projectile Motion
PHYSICS
• The Vertical (y) part of the motion:
vy = gt , y = (½)g
2
t
The SAME as free fall motion!!
 An object projected horizontally will reach
the ground at the same time as an object
dropped vertically from the same point!
(x & y motions are independent)
Projectile Motion
A common example
of a projectile is a
baseball!
A “Somewhat General” Case
An object is launched at initial angle θ0 with the horizontal. Analysis
of the motion is similar to before, except the initial velocity has a
vertical component vy0  0. Let up be positive now!
Components of initial velocity v0:
vx0 = v0cosθ0 vy0 = v0sinθ0
The Parabolic shape
of the path is real.
Acceleration = g down
for the entire trip!
• General Case: Take y positive upward &
origin at the point where it is shot: x0 = y0 = 0
vx0 = v0cosθ0, vy0 = v0sinθ0
• Horizontal Motion:
No Acceleration in the x Direction!
vx = vx0 , x = vx0 t
• Vertical Motion:
vy = vy0 - gt , y = vy0 t - (½)g t2
(vy) 2 = (vy0)2 - 2gy
• If y is positive downward, the - signs become + signs.
Summary: Projectile Motion
Projectile Motion
Motion with constant acceleration in 2 dimensions,
where the acceleration is g and is down.
Solving Projectile Motion Problems
1. Read the problem carefully, & choose the object(s) you
are going to analyze.
2. Sketch a diagram.
3. Choose an origin & a coordinate system.
4. Decide on the time interval; this is the same in both
directions, & includes only the time the object is
moving with constant acceleration g.
5. Solve for the x and y motions separately.
6. List known & unknown quantities. Remember that vx
never changes, & that vy = 0 at the highest point.
7. Plan how you will proceed. Use the appropriate
equations; you may have to combine some of them.
Projectile Motion
Example 4.4: Rolling off a Cliff
• A car rolls off a cliff of height
h = 20 m. Its initial velocity is
v0 = 10 m/s, along the
horizontal.
Calculate
a. The time it takes to hit the
ground at the base after it
leaves the cliff.
b. Its horizontal distance x from
the base when it hits the
ground.
c. Its velocity when it hits the
ground.
Example: Driving off a cliff!!
A movie stunt driver on a motorcycle speeds horizontally off a 50 m high
cliff. How fast must the motorcycle leave the cliff top to land on level
ground below, 90 m from the base of the cliff where the cameras are?
vx = vx0 = ? vy = -gt
x = vx0t, y = - (½)gt2
Time to bottom:
t = √2y/(-g) = 3.19 s
vx0 = (x/t) = 28.2 m/s
y is positive upward, y0 = 0 at top. Also vy0 = 0
Example: Kicked Football
lllllllll
• A football is kicked at an angle θ0 = 37.0° with a velocity of
20.0 m/s, as shown. Calculate: a. Max height. b. Time when
hits ground. c. Total distance traveled in the x direction. d.
Velocity at top. e. Acceleration at top.
θ0 = 37º, v0 = 20 m/s
 vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s