A baseball batter hits an incoming 40-m/s fastball.The ball leaves the bat at 50 m/s after a ball-on-bat contact time of 0.030 s.What is the force exerted on the 0.15-kg baseball?

Question

Quizplus · Accepted Answer

The force exerted on the baseball can be calculated using the impulse-momentum theorem: $ F \cdot \Delta t = m \cdot \Delta v $. Here, $ \Delta v = 50 \, \text{m/s} - (-40 \, \text{m/s}) = 90 \, \text{m/s} $. So, $ F = \frac{m \cdot \Delta v}{\Delta t} = \frac{0.15 \, \text{kg} \cdot 90 \, \text{m/s}}{0.030 \, \text{s}} = 450 \, \text{N} $.

A Baseball Batter Hits an Incoming 40-M/s Fastball