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 68.7 m across.  If he desires a 2.9-second flight time, what is the correct angle for his launch ramp (deg)?  

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 68.7 m across. If he desires a 2.9-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

h = vertical distance = -15 m
r = horizontal distance = 68.7 m
t = time = 2.9 s
g = acceleration by gravity = -9.8 m/s

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