a 0.055kg tennis ball approaches a tennis racket at 25m/s, stays in contact with it for 0.010s, is reversed in direction leaves the racket at 40m/s. what is the avg force between the racket and ball?

Answer 1

a=(V-Vo)/t = (-40-25)/0.01=-6500 m/s^2.
F = m*a = 0.055 * -6500 = -358 N.

Answer 2

a 0.005 kg tennis ball approaches a racket at an initial velocity 25 m/s2 . if its is in contact with the racket's strings for 0.005 s, then rebounds at 45 m/s2 what is the average contact force between the ball and the racjet?