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A batter hits a fastball with a mass of 0.145 kg that was traveling at 40.0 m/s; the force exerted by the bat on the ball is 29 N. What is the time of contact with the ball in order to reverse its motion at its original speed?

Answer :

LAMMETTHASHGO

The force applies an acceleration in the reverse direction of

29 N = (0.145 kg) a   →   a = 200 m/s²

and in order to hit the ball back with speed 40.0 m/s, this acceleration is applied over time t such that

200 m/s² = (40.0 m/s - (-40.0 m/s)) / t   →   t = (80.0 m/s) / (200 m/s²) = 0.4 s

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