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 62.5 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

x ? xo = (vox)t 69.4m = (vox)(3s) 23.13 m/s = vox Range =((vo^2)/g) sin(2?) 69.4m ={ (23.13^2 m/s)/9.81 m/s^2 } sin(2?) ? = error

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