V Vy Vy Vx V Projectile Motion Vx

V Vy Vy Vx. V Projectile Motion Vx

Horizontal Projectiles An applied horizontal force will cause an object to follow a curved path. After being launched, gravity pulls it downward.

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.

Horizontal Displacement: Where d = dx Use V= d/t d x = Vx t Vertical Displacement: Use d = vit + ½ at 2 vi = 0 m/s and a= 9. 81 m/s 2 dy dy = 1/2 gt 2 dx

Ex A) A stone is thrown horizontally from a cliff with a horizontal velocity of 15 m/s. If it takes 3. 0 seconds to hit the ground what is the height of the cliff? dy = 1/2 gt 2 dy = ½ 9. 81 m/s 2(3. 0 sec)2 dy= 44. 2 m

Ex B) How far from the base of the cliff does the stone fall? d x = Vx t dx = 15 m/s( 3. 0 sec) =44. 2 m = 45 m dx = 45 m

Your Turn • A soccer ball is kicked horizontally off a 22. 0 meter high hill and lands a distance of 35. 0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. • First find time in the air: dy=1/2 g t 2 22 m = ½ 9. 81 m/s 2 (t 2) t = 2. 12 seconds

• Use time to calculate the velocity: Vx = dx/t Vx = 35 m/2. 12 s Vx = 16. 51 m/s

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

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

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

Ex- A cannon ball is shot at an angle of 20 degrees with an initial velocity of 30 m/s. a) What are the vertical and horizontal components of the initial velocity? Vy = V sinѲ Vx = V cos Ѳ Vy = 30 m/s sin 20° Vx = 30 m/s cos 20° Vy = 10. 3 m/s Vx = 28. 2 m/s

b) What is the maximum vertical displacement of the cannon ball? Find peak time first: tpeak = Vy/g tpeak=10. 3 m/s/9. 81 m/s 2 tpeak = 1. 05 sec Now find dy: dy = 1/2 gt 2 dy = ½(9. 81 m/s 2)( 1. 05 sec)2 dy =5. 41 m

c) What horizontal distance does the cannon ball travel? Find total time first: t total = 2(tpeak) ttotal= 2(1. 05 sec) ttotal = 2. 1 sec Solve for dx: dx= Vxttotal dx = 28. 2 m/s (2. 1 sec) dx = 59. 22 m

Vy=10. 3 m/s Vx=28. 2 m/s dy=5. 41 m dx=59. 22 m