Answer:
The time it takes the ball to stop is 0.021 s.
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;
[tex]F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\[/tex]
Therefore, the time it takes the ball to stop is 0.021 s.
