A 41.6-kg boy, riding a 2.35-kg skateboard at a velocity of +5.82 m/s across a l

Question

A 41.6-kg boy, riding a 2.35-kg skateboard at a velocity of +5.82 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 5.40 m/s, 8.82° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Please Show work in a understandable fashion

Explanation / Answer

in horizontal direction

initial momentum=(41.6+2.35)*5.82=255.79

Final momentum=41.6*(5.4cos8.82)+2.35(vskateboard)=221.98+2.35(vskateboard)

initial momentum=Final momentum

255.79=221.98+2.35(vskateboard)

(vskateboard)=+14.39m/s relative to sidewalk

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