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Brandon is the catcher for a baseball team. He exerts a forward force on a 0.145 kg baseball to bring it to rest from a speed of 38.2 m/s. His hand recoils a distance of 0.135 m.

What is the acceleration of the ball and the force applied to it by Brandon?

Answer :

The force applied by Brandon to bring the ball to rest is 148.7 N.

We can use the equation F = ma to solve for the acceleration of the ball. Since the ball is being brought to rest from a velocity of 38.2 m/s, its initial velocity is 38.2 m/s and its final velocity is 0 m/s. Therefore, we have:

v² - u² = 2as

where v = final velocity, u = initial velocity, a = acceleration, and s = displacement. Solving for acceleration, we get:

a = (v² - u²) / (2s)

= (0 - (38.2 m/s)²) / (2 * -0.135 m)

= 1025.5 m/s² (rounded to 3 significant figures)

Therefore, the acceleration of the ball is 1025.5 m/s².

To find the force applied by Brandon, we can use the equation F = ma again, but this time we solve for force. Since the mass of the ball is 0.145 kg, we have:

F = ma

= 0.145 kg * 1025.5 m/s²

= 148.7 N (rounded to 3 significant figures)

Therefore, the force applied by Brandon to bring the ball to rest is 148.7 N.

Learn more about “ final velocity “ visit here;

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