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Aim: More projectile motion (trajectory) problems

Aim: More projectile motion (trajectory) problems. Do Now: A stuntman rides his motorcycle over a ramp with an initial vertical velocity of 22 m/s at an angle of 35° to the horizontal. Determine the following: Time of flight Maximum height achieved Total range. x y.

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Aim: More projectile motion (trajectory) problems

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  1. Aim: More projectile motion (trajectory) problems • Do Now: • A stuntman rides his motorcycle over a ramp with an initial vertical velocity of 22 m/s at an angle of 35° to the horizontal. Determine the following: • Time of flight • Maximum height achieved • Total range

  2. x y viy = 22 m/s a = -9.8 m/s2 d = 0 m t = ? t = ? d = vit + ½at2 0 m = (22 m/s)t + ½(-9.8 m/s2)t2 0 = 22t – 4.9t2 4.9t2 = 22t 4.9t = 22 t = 4.5 s

  3. x y viy = 22 m/s a = -9.8 m/s2 d = 0 m t = 4.5 s dmax = ? vfmax = 0 m/s tmax = 2.25 s t = 4.5 s dmax = viytmax + ½atmax2 dmax = (22 m/s)(2.25 s) + ½(-9.8 m/s2)(2.25 s)2 dmax = 24.7 m

  4. x y viy = 22 m/s a = -9.8 m/s2 d = 0 m t = 4.5 s vfmax = 0 m/s tmax = 1.55 s dmax = 24.7 m t = 4.5 s d = ? vix = ? vix = 31.4 m/s viy = 22 m/s 35° vix = ?

  5. x y viy = 22 m/s a = -9.8 m/s2 d = 0 m t = 4.5 s vfmax = 0 m/s tmax = 1.55 s dmax = 24.7 m t = 4.5 s vix = 31.4 m/s d = ? d = 141.3 m d = (31.4 m/s)(4.5 s) d = 141.3 m

  6. What direction was he accelerating at the maximum height? Down (gravity is the only acceleration and always points down) What direction was he accelerating at 1 s? Down (gravity is the only acceleration and always points down!) What will happen to his range if the ramp is raised to an incline of 45°? His range will increase (45° gives the largest range) What is his vertical velocity at the maximum height? 0 m/s What is his horizontal velocity at the maximum height? 31.4 m/s (horizontal velocity is constant)

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