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Projectile motion: Steps to solve popular problems.

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

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Projectile motion: Steps to solve popular problems.

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  1. Projectile motion:Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using dy = viyt + ½ at2 Note: viy = 0 Can find range using dx = vx 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?

  2. Projectile motion:Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using dy = viyt + ½ at2 Note: viy = 0 Can find range using dx = vx 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 = viyt + ½ at2 140 m = 0+ (.5*9.81*t2) t = 5.34 s

  3. Projectile motion:Steps to solve popular problems. Scenario 1a: Given velocity shot horizontally off a cliff with a height. Can find time using dy = viyt + ½ at2 Note: viy = 0 Can find range using dx = vx 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 = viyt + ½ at2 140 m = 0+ (.5*9.81*t2) t = 5.34 s dx = vx t dx = (15m/s)(5.34s) dx = 80.14 m

  4. 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?

  5. 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 = viyt + ½ at2 • 1.3m = 0+ (.5*9.81*t2) • t = .515 s

  6. 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 = viyt + ½ at2 • 1.3m = 0+ (.5*9.81*t2) • t = .515 s • dx = vx t • dx = (60m/s)(.515s) • dx = 30.9 m

  7. Scenario 2a: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note:viy = 0 • Can find range using dx = vx 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?

  8. Scenario 2a: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note:viy = 0 • Can find range using dx = vx 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 = viyt + ½ at2 • dy = 0+ (.5)(9.81m/s2)(2.6s)2 • dy = 33.16m

  9. Scenario 2a: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note: viy = 0 • Can find range using dx = vx 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 = viyt + ½ at2 • dy = 0+ (.5)(9.81m/s2)(2.6s)2 • dy = 33.16m • dx = vx t • dx = (50 m/s)(2.6s) • dx = 130 m

  10. Scenario 2b: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note: viy = 0 • Can find range using dx = vx 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?

  11. Scenario 2b: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note: viy = 0 • Can find range using dx = vx 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 = viyt + ½ at2 • dy = 0+ (.5*9.81m/s2*1.5s2) • dy = 11m

  12. Scenario 2b: • Given velocity shot horizontally off a cliff with time in air. • Can find height using dy = viyt + ½ at2Note: viy = 0 • Can find range using dx = vx 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 = viyt + ½ at2 • dy = 0+ (.5*9.81m/s2*1.5s2) • dy = 11m • dx = vx t • dx = (16 m/s)(1.5s) • dx = 24 m

  13. Scenario 3a: Given projectile shot with a velocity and angle. • Find components • Vx = V Cos θ , Vy = V Sin θ • Find time to top Vf = Viy + at • Vf = 0, a = -9.81m/s2 • Find height using any of the following formulas: • dy = viyt + ½ at2 a = -9.81 m/s2 • Vf2 = Viy2 + 2 a dy • Ave V = d/t Ave V is ½ of Viy • Find time down use the adjusted height with dy = viyt + ½ at2 :viy = 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 (dx) dx = Vx ttotal • A golfer hits a ball at 60 m/s at 22 degrees.

  14. 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 = 55.63 m/s • Vy = 60 Sin 22 • Vy = 22.48 m/s

  15. Find time to topVf = Viy + at Vf = 0, a = -9.81m/s2 • Vf = Viy + at • 0 m/s = 22.48 m/s + (-9.81m/s2)(t) • - 22.48 m/s = (-9.81m/s2)(t) • ttop = 2.29 s

  16. Find height using any of the following formulas:dy = viyt + ½ at2 a = -9.81 m/s2Vf2 = Viy2 + 2 a dyAve V = d/t Ave V is ½ of Viy Vf2 = Viy2 + 2 a dy (0 m/s)2 = (22.48 m/s)2 + (2)(-9.81 m/s2)(dy) dy = 25.76 m

  17. Find time down use the adjusted height with dy = viyt + ½ at2Note: viy = 0 dy = viyt + ½ at2 25.76 = 0 + ½ 9.81 m/s2 t2 tdown = 2.29s

  18. 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 (dx) dx = vx ttotal tup = 2.29s tdown = 2.29s ttotal = 4.58s dx = vx ttotal dx = (55.63 m/s)(4.58s) dx = 254.79 m

  19. 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

  20. A punter kicks a football at 33 m/s at and angle of 57 degrees. • Find: • X and Y Components • Vx = 17.97 m/s • Vy = 27.68 m/s • Time up • tup = 2.82 s • Height • 39.05 m • Total Time in air. • 5.64 s • Range • dx = 101.35m

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