Projectile motion: Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using d y = vi y t + ½ at 2 Note: vi y = 0 Can find range using d x = v x t A bird travelling at 15 m/s drops a rock to the ground 140 meters below. How far horizontally from initial distance did the rock land?

Projectile motion: Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using d y = vi y t + ½ at 2 Note: vi y = 0 Can find range using d x = v x t A bird travelling at 15 m/s drops a rock to the ground 140 meters below. How far horizontally from initial distance did the rock land? dy = vi y t + ½ at m = 0+ (.5*9.81*t 2 ) t = 5.34 s

Projectile motion: Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using d y = vi y t + ½ at 2 Note: vi y = 0 Can find range using d x = v x t A bird travelling at 15 m/s drops a rock to the ground 140 meters below. How far horizontally from initial distance did the rock land? dy = vi y t + ½ at m = 0+ (.5*9.81*t 2 ) t = 5.34 s d x = v x t d x = (15m/s)(5.34s) d x = m

Scenario 1b: A boy shoots a rock horizontally from his slingshot. The rock is released 1.3 meters from ground level. The rock is travelling at 60 m/s. How far did the rock go?

Scenario 1b: A boy shoots a rock horizontally from his slingshot. The rock is released 1.3 meters from ground level. The rock is travelling at 60 m/s. How far did the rock go? dy = vi y t + ½ at 2 1.3m = 0+ (.5*9.81*t 2 ) t =.515 s

Scenario 1b: A boy shoots a rock horizontally from his slingshot. The rock is released 1.3 meters from ground level. The rock is travelling at 60 m/s. How far did the rock go? dy = vi y t + ½ at 2 1.3m = 0+ (.5*9.81*t 2 ) t =.515 s d x = v x t d x = (60m/s)(.515s) d x = 30.9 m

Scenario 2a: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note:vi y = 0 –Can find range using d x = v x t A car travelling at 50 m/s slides off the road and lands in a creek. The car was in the air for 2.6 seconds. How far down and how far from the edge of the road did the car land?

Scenario 2a: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note:vi y = 0 –Can find range using d x = v x t A car travelling at 50 m/s slides off the road and lands in a creek. The car was in the air for 2.6 seconds. How far down and how far from the edge of the road did the car land? dy = vi y t + ½ at 2 d y = 0+ (.5)(9.81m/s 2 )(2.6s) 2 d y = 33.16m

Scenario 2a: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note: vi y = 0 –Can find range using d x = v x t A car travelling at 50 m/s slides off the road and lands in a creek. The car was in the air for 2.6 seconds. How far down and how far from the edge of the road did the car land? dy = vi y t + ½ at 2 d y = 0+ (.5)(9.81m/s 2 )(2.6s) 2 d y = 33.16m d x = v x t d x = (50 m/s)(2.6s) d x = 130 m

Scenario 2b: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note: vi y = 0 –Can find range using d x = v x t  John throws a rock horizontally at 16 m/s off a bridge. The rock hits the ground 1.5 seconds later how high is John and how far did the rock travel?

Scenario 2b: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note: vi y = 0 –Can find range using d x = v x t  John throws a rock horizontally at 16 m/s off a bridge. The rock hits the ground 1.5 seconds later how high is John and how far did the rock travel? dy = vi y t + ½ at 2 d y = 0+ (.5*9.81m/s 2 *1.5s 2 ) d y = 11m

Scenario 2b: Given velocity shot horizontally off a cliff with time in air. –Can find height using d y = vi y t + ½ at 2 Note: vi y = 0 –Can find range using d x = v x t  John throws a rock horizontally at 16 m/s off a bridge. The rock hits the ground 1.5 seconds later how high is John and how far did the rock travel? dy = vi y t + ½ at 2 d y = 0+ (.5*9.81m/s 2 *1.5s 2 ) d y = 11m d x = v x t d x = (16 m/s)(1.5s) d x = 24 m

Scenario 3a: Given projectile shot with a velocity and angle. –Find components V x = V Cos θ, V y = V Sin θ –Find time to top V f = V iy + at V f = 0, a = -9.81m/s 2 –Find height using any of the following formulas: d y = v iy t + ½ at 2 a = m/s 2 V f 2 = V iy a d y Ave V = d/tAve V is ½ of V iy –Find time down use the adjusted height with d y = v iy t + ½ at 2 :v iy = 0 –Total Time add the time up to time down. If there is no change in elevation of the target then just double the time up. –Range (d x ) d x = V x t total A golfer hits a ball at 60 m/s at 22 degrees.

Find components Vx = V Cos θ, Vy = V Sin θ Vx = 60 Cos 22 Vx = m/s Vy = 60 Sin 22 Vy = m/s

Find time to top V f = V iy + at V f = 0, a = -9.81m/s 2 V f = V iy + at 0 m/s = m/s + (-9.81m/s 2 )(t) m/s = (-9.81m/s 2 )(t) t top = 2.29 s

Find height using any of the following formulas: d y = vi y t + ½ at 2 a = m/s 2 V f 2 = Vi y a d y Ave V = d/t Ave V is ½ of V iy V f 2 = Vi y a d y (0 m/s) 2 = (22.48 m/s) 2 + (2)(-9.81 m/s 2) (d y ) d y = m

Find time down use the adjusted height with d y = v iy t + ½ at 2 Note: v iy = 0 d y = v iy t + ½ at = 0 + ½ 9.81 m/s 2 t 2 t down = 2.29s

Total Time add the time up to time down. If there is no change in elevation of the target then just double the time up. Range (d x ) d x = v x t total t up = 2.29s t down = 2.29s t total = 4.58s d x = v x t total d x = (55.63 m/s)(4.58s) d x = m

A punter kicks a football at 33 m/s at and angle of 57 degrees. Find: X and Y Components Time up Height Total Time in air. Range

A punter kicks a football at 33 m/s at and angle of 57 degrees. Find: X and Y Components –V x = m/s –V y = m/s Time up –t up = 2.82 s Height –39.05 m Total Time in air. –5.64 s Range d x = m