A 38.0-kg boy, riding a 2.37-kg skateboard at a velocity of +5.07 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 6.06 m/s, 10.35° 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.

Answer 1

x momentum after = x momentum before

after
p = 38 (6.06 cos 10.5) + 2.37 v

before
p = (38+2.37)(5.07)

set equal, solve for v which will probably be negative (kid goes forward, board goes backward)

Answer 2

Damon, Thank you so much