Answer to Question #144028 in Physics for Katy R Lopez

Question

A baseball is the thrown at 30 m/s. The baseball has mass of 0.5 kg. The ball was in the hand of the pitcher for 0.3 seconds. How large a force did the pitcher exert on the ball?

Answer 1

We can find the average force that the pitcher exert on the ball from the impulse-momentum change equation:

$$F_{avg}\Delta t=m\Delta v=m(v_f-v_i),$$

here, $F_{avg}$ is the average force that the pitcher exert on the ball, $\Delta t=0.3\ s$ is the time during that the ball was in the hand of the pitcher, $m=0.5\ kg$ is the mass of the ball, $\Delta v$ is the change in the velocity of the ball, $v_i=30\ \dfrac{m}{s}$ is the initial velocity of the ball, $v_f=0\ \dfrac{m}{s}$ is the final velocity of the ball.

Then, from this equation we can find the magnitude of the average force that the pitcher exert on the ball:

$$F_{avg}=\dfrac{m \lvert (v_f-v_i) \rvert}{\Delta t},$$ $$F_{avg}=\dfrac{0.5\ kg\cdot \lvert (0\ \dfrac{m}{s}-30\ \dfrac{m}{s}) \rvert}{0.3\ s}=50\ N.$$

Answer:

$F_{avg}=50\ N.$