1. A projectile is any object that moves through
Projectile Motion!
A projectile is any object that moves through
the air or through space, acted on only by gravity (and air resistance).

2. Projectile motion is divided into two sections:
Objects launched horizontally
Objects launched at angles

3. HORIZONTAL PROJECTION
Motion in which something is launched horizontally.
X component remains Constant.
Vertical Motion of a
Projectile launched
Horizontally
Horizontal Motion

4. HORIZONTAL PROJECTION
Horizontal velocity is constant. Vertical velocity is changing due to gravitational acceleration..

5. When solving projectile motion problems it is important to separate motion along the x axis and y axis. Time is universal, it will be the same for both x and y.
X Y
Vox= Voy =
Vx = Vy =
ΔX = ΔY=
a = a=
t = t=

6. y = -1/2at2 y = -1/2(9.8)(5)2 y = -1/2(9.8)(5)2 y = -122.5 m
Example
A ball is thrown horizontally at a speed of 30 m/s from the top of a cliff. If the ball hits the ground 5.0 seconds later, approximately how high is the cliff?
Vo= 30 m/s
y = -1/2at2 g y
y = -1/2(9.8)(5)2
y = -1/2(9.8)(5)2
t= 5s
y = m
y = m

7. Take out a separate sheet of paper. We are going to do #3 together!

8. Horizontal Projectile Recap:
We need to split motion into x-component and y-component
With Horizontal Projectiles, the Vi in only in the x direction and the acceleration is only in the y-direction. We can simplify the formulas:
Displacement Equations:
ΔX = Vi,xt + ½ axt2
ΔY = Vi,yt + ½ ayt2
ax = 0
Vi,y = 0
ΔX = Vi,xt
ΔY = ½ ayt2
Time can help you solve for a missing variable!

9. Horizontal Projectile Recap:
Horizontal speed stays constant. Vertical velocity will change due to gravity.
Velocity Equations:
Vi,x = Constant
Vf,y = Vi,y + ayt
Vf,y2 = Vi,y2 + 2ayΔy
Vf,y = ayt
Vf,y2 = 2ayΔy
There is no Vi,y

10. Why is this important? What are the real world applications?

11. PROJECTILES AT AN ANGLE
The more general case of projectile motion occurs when the projectile is fired at an angle.

12. PROJECTILES AT AN ANGLE
At what angle will the projectile will travel the furthest?
At what angle will the projectile stay in the air longer?
75°
(at 90° you get max time in the air!)
45°

13. PROJECTILES AT AN ANGLE
The horizontal velocity component remains the same size throughout the entire motion of the cannonball.

15. Projectiles Launched at Angles Recap:
We need to split motion into x-component and y-component Vy
Vx = Vo cosθ
Vy = Vo sinθ
10 m/s
Vx = 10 cos(30)
Vy = 10 sin(30)
30° Vx
Vx = 8.66 m/s
Vy = 5 m/s
ΔX = Vi,xt + ½ axt2
ΔY = Vi,yt + ½ ayt2
ax = 0
Time can help you solve for a missing variable!

16. Projectiles launched at anglesRecap:
Horizontal speed stays constant. Vertical velocity will change due to gravity.
First, break the initial velocity into components.
Vx = Vo cosθ
Vy = Vo sinθ
Velocity Equations:
Vi,x = Constant
Vf,y = Vi,y + ayt
Vf,y2 = Vi,y2 + 2ayΔy

17. Assumptions you CAN make.
Acceleration in the y- direction will be -9.8 m/s2 At maximum height Vy= 0 m/s Velocity in the x direction in constant Air resistance is ignored (unless otherwise stated) If a projectile is launched at an angle, it has Vx and Vy If a projectile is launched horizontally, Vy=0

18. Vx = Vo cosθ Vy = Vo sinθ Vx = 200 cos40 Vy = 200 sin40
X Y
Vix=
m/s
Viy=
128.56m/s
ax= 0 m/s2
ay= -9.8 m/s2
Vx = 200 cos40
Vy = 200 sin40
Δx=
Δy=
Vfx=
153.21m/s
Vfy=
m/s t= t=
Vx = m/s
Vy = m/s
200 m/s Vy
40° Vx
-Vfy = Viy

19. Vfy = Viy + at -128.56 = 0 + (-9.8)t 26.24= t -Vfy = Viy 200 m/s X Y
Vix=
m/s
Viy=
128.56m/s
ax= 0 m/s2
ay= -9.8 m/s2
= 0 + (-9.8)t
Δx=
Δy=
26.24= t
Vfx=
m/s
Vfy=
m/s
26.24s t= t=
26.24s
200 m/s Vy
40° Vx
-Vfy = Viy

20. Vix = Δx t Δx = Vix *t Δx = 153.21 *26.24 Δx = 4020.23 -Vfy = Viy
X Y
Vix=
153.21
Viy=
128.56
ax= 0
ay= -9.8
Δx = Vix *t
Δx=
Δy=
Vfx=
153.21
Vfy=
Δx = *26.24 t=
26.24s t=
26.24s
Δx =
200 m/s Vy
40° Vx
-Vfy = Viy

21. Vfy2 = Viy2 + 2aΔy 02 = 128.562 + 2(-9.8)Δy -128.562 = 2(-9.8)Δy
X Y
Vfy2 = Viy2 + 2aΔy
Vix=
153.21
Viy=
128.56
02 = (-9.8)Δy
ax= 0
ay= -9.8
Δx=
Δy=
843.24
= 2(-9.8)Δy
Vfx=
153.21
Vfy=
843.24= Δy t=
26.24s t=
26.24s
max height = 0
200 m/s Vy
40° Vx
-Vfy = Viy