A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcyc

Question

A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 61.7 m across. If he desires a 3.0-second flight time, what is the correct angle for his launch ramp (deg)?What is his correct launch speed?
What is the correct angle for his landing ramp (give a positive angle below the horizontal)? What is his predicted landing velocity. (Neglect air resistance.)

Explanation / Answer

y = 15.0 m, x = 61.7 m, t = 3.0-s Find angle A (below horizontal) and launch speed v x = vcos(A)*t, so vcos(A) = x/t = 20.57 m/s y = vsin(A)*t - gt^2/2, so vsin(A) = y/t - gt/2 = -9.70 m/s v = sqrt(20.57^2 + 9.70^2) = 22.7 m/s A = arcsin(-9.70/22.7) = -25.2 degrees

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