Answer:
The time it takes the ball to stop is 0.021 s.
Step-by-step explanation:
Given;
mass of the softball, m = 720 g = 0.72 kg
velocity of the ball, v = 15.0 m/s
applied force, F = 520 N
Apply Newton's second law of motion, to determine the time it takes the ball to stop;
![F = ma = (mv)/(t) \\\\t = (mv)/(F) \\\\t = (0.72 \ * \ 15)/(520) \\\\t = 0.021 \ s \\](https://img.qammunity.org/2022/formulas/physics/college/3edgysngqqqpajjdw4tr1efb0bg8ylheqr.png)
Therefore, the time it takes the ball to stop is 0.021 s.