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. (See the figure.) The takeoff ramp was inclined at = 54.0 degrees, the river was L= 33.0m wide, and the far bank was d = 15.0m lower than the top of the ramp. The river itself was H= 100m below the ramp. You can ignore air resistance 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? Express your answer with the appropriate units. If his speed was only half the value found in part A, where did he land? Please go step by step because the calculations for the solution was not making sense.

Explanation / Answer

y component of velocity =vsin54 =0.81v x component of veocity =vcos54 =0.58v s =ut +1/2 at^2 15 =0.81vt -4.9t^2 hence ie 4.9t^2-0.81vt+15 =0 t =[0.81v+root(0.66v^2-294)]/(2*4.9) =[0.81v+root(0.66v^2-294)]/9.8.....1 also 0.58vt =33 hence t =56.9/v......2 equating 1 and 2 u will get the value of v ........... hope this will help :-)

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