A 0.15 kg baseball moving at +21.20 m/s is slowed to a stop by a catcher who exerts a constant force of -362 N.How long does it take this force to stop the ball? sHow far does the ball travel before stopping?

To solve this problem, we will use the second law of motion (F = ma) and the equations of motion.

Step 1: Find the acceleration of the baseball.

We know that F = ma, where F is the force, m is the mass, and a is the acceleration. We can rearrange this equation to solve for a:

a = F/m

Substituting the given values:

a = -362 N / 0.15 kg = -2413.33 m/s²

The negative sign indicates that the acceleration is in the opposite direction of the motion.

Step 2: Find the time it takes to stop the ball.

We know that final velocity (v) = initial velocity (u) + acceleration (a) * time (t). Since the ball is stopped, the final velocity is 0. We can rearrange this equation to solve for t:

t = (v - u) / a

Substituting the given values:

t = (0 - 21.20 m/s) / -2413.33 m/s² = 0.00878 s

Step 3: Find the distance the ball travels before stopping.

We know that distance (s) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * time (t)². Substituting the given values:

s = 21.20 m/s * 0.00878 s + 0.5 * -2413.33 m/s² * (0.00878 s)² = 0.093 m

So, it takes approximately 0.00878 seconds to stop the ball, and the ball travels approximately 0.093 meters before stopping.