1. P R O J E C T I L s V Vy Vx

2. Horizontal Projectiles
An applied horizontal force will cause an object to follow a curved path. After being launched the only force acting on it is gravity so it will hit the ground at the same time as an object in free fall.

3. As an object falls the horizontal velocity remains constant while the vertical velocity increases due to gravity.
If the curve of the projectile matches the curve of the earth, the projectile becomes a satellite.

4. Horizontal Displacement: Use V= d/t Where d = dx dx = Vx t
Vertical Displacement: Use d = vit + ½ at2
vi = 0 m/s and a= 9.81m/s2
dy = 1/2gt2 dy dx

5. Ex A) A stone is thrown horizontally from a cliff with a horizontal velocity of 15m/s. If it takes 3.0 seconds to hit the ground what is the height of the cliff?
dy = 1/2gt2
dy = ½ 9.81m/s2(3.0sec)2
dy= 44.2m

6. dx = Vx t dx = 15m/s( 3.0 sec) dx = 45m
Ex B) How far from the base of the cliff does the stone fall?
dx = Vx t
dx = 15m/s( 3.0 sec)
dx = 45m
=44.2m
= 45m

7. PROJECTILES AT AN ANGLE

8. At zero degrees, the projectile acts as a horizontal projectile.
At 45 degrees, the projectile will have the greatest horizontal displacement.
At 90 degrees the projectile will have the greatest vertical displacement but no horizontal displacement.

9. The horizontal velocity will help determine the horizontal displacement.
Vx = V cos Ѳ
The vertical component of the velocity will help determine how long the projectile is in the air and how high up it travels.
Vy = V sin Ѳ

10. Horizontal displacement: (use total time in air)
Vertical displacement: (use peak time)
dy = 1/2gt2
Time : Use a=ΔV/t a = g ΔV = Vy- 0m/s
tpeak = Vy/g ttotal = 2(tpeak)
dx = Vx ttotal

11. Ex- A cannon ball is shot at an angle of 20 degrees with an initial velocity of 30m/s.
a) What are the vertical and horizontal components of the initial velocity?
Vy = V sinѲ
Vy = 30m/s sin 20°
Vy = 10.3m/s
Vx = V cos Ѳ
Vx = 30m/s cos 20°
Vx = 28.2m/s

12. b) What is the maximum vertical displacement of the cannon ball?
Find peak time first:
tpeak = Vy/g tpeak=10.3m/s/9.81m/s2
tpeak = 1.05 sec
Now find dy: dy = 1/2gt2
dy = ½(9.81m/s2)( 1.05sec)2
dy =5.41m

13. c) What horizontal distance does the cannon ball travel?
Find total time first: t total = 2(tpeak) ttotal= 2(1.05sec) ttotal = 2.1sec Solve for dx: dx= Vxttotal dx = 28.2m/s (2.1sec) dx = 59.22m

14. Vy=10.3m/s
dy=5.41m
dx=59.22m
Vx=28.2m/s

15. The zookeeper would like to throw a banana to the monkey for a snack
The zookeeper would like to throw a banana to the monkey for a snack. If the monkey falls from the tree at the exact time the zookeeper throws the banana should he aim at the monkey, slightly above the monkey or slightly below the monkey?