Answer to Question #171881 in Atomic and Nuclear Physics for Nema

Question

What was the force being applied by the pitcher's hand if the ball is in contact with it for 0.2s?

Answer 1

Let's assume that the mass of the ball is $m = 150g = 0.15kg$, and the final speed is $v_f = 40m/s$.

Then the accleration of the ball during the contact is:

$$a = \dfrac{v_f - v_i}{t}$$

where $v_i = 0m/s$ is the initial speed of the ball, and $t = 0.2s$ is the time of the contact. Thus, obtain:

$$a = \dfrac{40m/s}{0.2s} = 200 m/s^2$$

According to the second Newton's law, the force applied by the pitcher's hand is:

$$F = ma\\ F = 0.15kg\cdot 200 m/s^2 = 30N$$

Answer. 30N.