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

Question

2 years ago

13

N. What is the time of contact with the ball in order to reverse its motion at its original speed?

1 answer:

Vika [28.1K] · Answer 1 · 2022-12-12T14:45:38+03:00

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