Question

A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s (103 mi/h). Determine the average force exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)

Answer 1

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

$F=ma$

a is acceleration

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N$

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.