Projectile Motion Thank you Physics Classroom: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html

Projectiles Bullet Fired Objects that only have the force of gravity acting on them. Have two parts: Horizontal Vertical Calculate them seperately

Paths of Projectiles

SOLVE: If the cannon ball left the cannon with a velocity of 20m/s SOLVE: If the cannon ball left the cannon with a velocity of 20m/s. Determine vertical and horizontal displacement at six seconds. How would you determine vertical distance at 6 seconds? dy = ½ a t2 176 m How would you determine horizontal distance at 6 seconds? v = d/t or dx = vt 120 m

2. A ball is thrown horizontally at 25 m/s off a roof 15 m high 2. A ball is thrown horizontally at 25 m/s off a roof 15 m high. How far does the ball travel before hitting the ground. First: Set up 2 columns for vertical and horizontal knowns Second: Determine how long the ball is in the air using vertical information Third: Use calculated time to determine horizontal distance.

Set up knowns for vertical and horizontal 3. A bullet is shot horizontally at 255 m/s. The gun barrel is 1.75 meters off the ground. How long before the bullet hits the ground? What distance will the bullet travel? What is the final vertical velocity? What is the final horizontal velocity? Set up knowns for vertical and horizontal

Easy A cannonball is shot straight off of a 20 m high cliff at 50 m/s Easy A cannonball is shot straight off of a 20 m high cliff at 50 m/s. Determine the range of the cannonball, the time the cannonball was in the air and the speed at which the cannonball hit the ground. t= 2dy g vx = 50 m/s t= 2(20) 9.8 t= 2.02 dy 20 m dx = 101m Horizontal Constant never changes. Vertical Changes due to gravity. vx dy = 20 m = 50 m/s t= 2.02 = 0 m/s always! When fired parallel with the ground viy dx = 50 (2.02) = 101 m T=2.02 a = -9.8 m/s 2 vfy = gt Vfy = -9.8 (2.02) = -19.79 m/s

More on Horizontal velocity More on Horizontal velocity horizontal velocity = tricky test questions An unfortunate accident occurred on the tollway. A driver accidentally passed through a faulty barricade on a bridge. and landed in a pile of hay. Measurements at the accident scene reveal that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. If the car was going 60 m/s when it went off of the bridge what was it’s final horizontal velocity?

2. A ball is thrown horizontally at 25 m/s off a roof 15 m high 2. A ball is thrown horizontally at 25 m/s off a roof 15 m high. How far does the ball travel before hitting the ground. First: Set up 2 columns for vertical and horizontal knowns Second: Determine how long the ball is in the air using vertical information Third: Use calculated time to determine horizontal distance.

Set up knowns for vertical and horizontal 3. A bullet is shot horizontally at 255 m/s. The gun barrel is 1.75 meters off the ground. How long before the bullet hits the ground? What distance will the bullet travel? What is the final vertical velocity? What is the final horizontal velocity? Set up knowns for vertical and horizontal

What happens if ball is launched straight up into the air? You throw a ball straight up into the air at 15m/s. What happens to it’s velocity as it rises? It slows What happens to it’s acceleration? It stays the same!!!

What happens if ball is launched straight up into the air? You throw a ball straight up into the air at 15m/s. What happens to it’s velocity as it falls? It increases What happens to it’s acceleration? It stays the same!!!

Ex. 1 An eight ball rolls at1. 5 m/s off of a 1. 2 m – high table Ex.1 An eight ball rolls at1.5 m/s off of a 1.2 m – high table. What is the balls range when it strikes the ground? Horizontal Vertical Constant never changes. Changes due to gravity. vx dy viy dx vfy a = -9.8 m/s 2

Steps to solving a projectile problem Horizontal Vertical Steps to solving a projectile problem Launched at an angle? Straight at an angle straight at an angle NO YES Vx = dx t viy= 0 a=-9.8 vfy = gt 1st Find time Viy=0 a=-9.8 dx = Vxt 1st Find Viy and Vx using trig dy = gt2 2 2nd Find time t= 2dy g g = -9.8 ttotal = Vfy - Viy t= 2dy g g

HARD A cannonball is shot at 50 m/s 58° above the horizontal HARD A cannonball is shot at 50 m/s 58° above the horizontal. Determine the range of the cannonball, the maximum height, and the time the cannonball was in the air. Viy V=50 m/s dy Vx dx Horizontal Vertical Constant never changes. Changes due to gravity. Vx =   26.5 m/s Viy =   42.4 m/s dx - 42.4 m/s Vfy = -Viy = ttotal = Vfy - Viy dy g

Ex. 1 Horizontal Vertical dx dy= g Vx = Viy = Vfy = Vfy - Viy ttotal = Constant never changes. Changes due to gravity. Vx = Viy = dx Vfy = dy= ttotal = Vfy - Viy g

Steps to solving a projectile problem Launched at an angle? NO YES a=-9.8 1st Find time Viy=0 a=-9.8 1st Find Viy and Vx using trig 2nd Find time t= 2dy g ttotal = Vfy - Viy g

Projectiles. From Physclips: Mechanics with animations and film.