A physics professor did daredevil stunts in his spare time. His last stunt was a

Question

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. The takeoff ramp was inclined at alpha = 54.0 degrees, the river was L = 40.0 m wide, and the far bank was d = 20.0 m lower than the top of the ramp. The river itself was H = 77.0 { m m} below the ramp. You can ignore air resistance.

a. What should his speed have been at the top of the ramp for him to have just made it to the edge of the far bank? answer is 17.4 m/s

b. If his speed was only half the value found in part A, where did he land?

Explanation / Answer

angle A = 53.0 degrees, sinA = 0.8, cosA = 0.6 d = 40.0 m, h = 15.0 m, H = 100 m, (a) find speed v vertical displacement -h = vsinA*t - gt^2/2 -15 = 0.8vt - 4.9t^2 (1) horizontal displacement d = vcosA*t = 0.6vt so vt = d/0.6 = 40/0.6 plug it into (1) and get -15 = 0.8*40/0.6 - 4.9t^2 t = 3.734 s v = (40/0.6)/t = 17.8 m/s (b) If his speed was only half the value found in (a), where did he land? v = 17.8/2 = 8.9 m/s vertical displacement = -H = vsinA*t - gt^2/2 4.9t^2 - 8.9*0.8t - 100 = 0 t = 5.30 s d = vcosA*t = 28.3 m

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