1. PROJECTILE MOTION!
• Recall…
• In projectile motion, the
vertical and horizontal
components are independent
from one another & must be
analyzed that way.
• Velocity in the x direction is
constant
• Velocity in the y-direction is
influenced by the force of
gravity

2. 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

3. VERTICALLY LAUNCHED PROJECTILES
Since the projectile was launched at a angle, the velocity
MUST be broken into components!!!
Vxi = Vi cosθ
vi vyi
Vyi = Visinθ
q
vxi

4. 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

5. VERTICALLY LAUNCHED PROJECTILES
You will still use kinematic equations, but YOU MUST use
COMPONENTS in the equation.
vi vyi X = Vxi t yf = yi + Vyit+0.5gt2
q
vxi Vxi = Vi cosθ
Vyi = Visinθ

6. 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?
4 m/ s Step 1: Break initial velocity into
(c) How high does it travel?
Its vector components
7m/ s Vxi = Vi cosθ
Vxi = 20cos53 = 12.04 m/s
q  53 Vyi = Visinθ
Vyi = 20sin53 = 15.97 m/s

7. EXAMPLE X y
A place kicker kicks a xi 0m 0m
football with a velocity of xf ? 0m
20.0 m/s and at an angle
of 53 degrees. vi Vxi = 12.04 m/s Vyi =15.97 m/s
(a) How long is the ball in
the air? vf
a 0 - 9.8 m/s/s
yf = yi + Vyit+0.5gt2 t ?
0 = 0 + 15.97t + 0.5 (-9.8)t2 
-15.97t = -4.9t2  -15.97 = -4.9 t  t = 3.26 s

8. X y
EXAMPLE xi 0m 0m
A place kicker kicks a xf ? 0m
football with a velocity of vi Vxi = 12.04 m/s Vyi =15.97
20.0 m/s and at an angle m/s
of 53 degrees.
(b) How far away does it vf
land? a 0 - 9.8 m/s/s
t 3.26s
 X = vxit  (12.04)(3.26) = 39.24 m

9. X y
EXAMPLE xi 0m 0m
xf ? ?
A place kicker kicks a football
with a velocity of 20.0 m/s and vi Vxi = 12.04 m/s Vyi =15.97 m/s
at an angle of 53 degrees. vf
a 0 - 9.8 m/s/s
(c) How high does it travel? t 3.26s ÷ 2 = 1.63 s
CUT YOUR TIME IN HALF!
yf = yi + Vyit+0.5gt2
yf = 0 + 15.97(1.26) + 0.5 (-9.8)(1.26)2 = 13.01 m

10. EXAMPLE: X y
xi 0m 0m
A soccer ball is kicked towards xf ? 0m
the goal at 12 m/s at an angle
vi 9.83 m/s 6.88 m/s
of 35° from the ground, and
just makes it into the goal. vf
How far away was the player a 0 - 9.8 m/s/s
from the goal when they t 1.40 s
kicked the ball?
1. Write down all information & X = 13.8 m
draw a diagram
2. Break Vi into its components
3. Use equations to solve for x.

11. PROJECTILE MOTION WORKSHEET

12. TRAJECTORY AND RANGE
• Maximum range is at 45°
• Low and high trajectory
cover the same distance.
• 30 and 60
• 10 and 80
• 25 and…

14. THE RANGE EQUATION!
• Range: Maximum horizontal distance of a projectile.
• Equation is derived on page 64 in your book…
X = R = 2Vo2sinθcosθ
g
Using the trig substitution
2sinθcosθ = sin2θ, we get…
X = R = Vo2sin2θ
g Vyi
θ
Vxi
R

15. RANGE
• When solving for theta (θ) using the range equation, you must
solve for both complementary angles
Complementary angles = θ or 90-θ