A 39.2-kg boy, riding a 2.39-kg skateboard at a velocity of +4.97 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 6.48 m/s, 9.54° 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?
2 answers
V = 6.48*cos9.54 = 6.39 m/s.
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