At the instant a race began, a 60-kg sprinter exerted a force of 669-N on the starting block at a 24 angle with respect to the ground. If this force was exerted for 0.44-s, with what speed (in m/s) did the sprinter leave the starting block?

Answer 1

Fx = 669 cos 24

change in x momentum or impulse = Fx t
so

669 cos 24 * .44 = 60 v