A baseball is pitched at 40 m/s. The batter hits the ball directly back at the pitcher at 50 m/s. What is the magnitude of the acceleration of the ball if it was in contact with the bat for 1/30s. Answer says 2700m/s^2

Question

A baseball is pitched at 40 m/s. The batter hits the ball directly back at the pitcher at 50 m/s. What is the magnitude of the acceleration of the ball if it was in contact with the bat for 1/30s. Answer says 2700m/s^2

Answer 1

To calculate the magnitude of acceleration, we can use the equation:

acceleration = (change in velocity) / (time taken)

First, let's find the change in velocity. The ball's initial velocity is 40 m/s when pitched and its final velocity is -50 m/s when hit back by the batter. Since the final velocity is in the opposite direction, we take -50 m/s as the change in velocity:

change in velocity = -50 m/s - 40 m/s = -90 m/s

Next, we need to find the time taken. The problem states that the ball was in contact with the bat for 1/30 seconds:

time taken = 1/30 s

Now we can calculate the acceleration:

acceleration = (-90 m/s) / (1/30 s)
= (-90 m/s) * (30 s)
= -2700 m/s^2

Therefore, the magnitude of the acceleration of the ball is 2700 m/s^2. Since acceleration is a vector quantity, the negative sign indicates that the acceleration is in the opposite direction to the initial velocity.