Question

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51m/s to the north. What was the average acceleration of the ball during the 1.0ms when it was in contact with the bat?

Answer 1

Let us assume the south direction as positive and the North direction as negative.

Acceleration at the point of contact
= (43-(-51))/0.001 s
= 94000m/s²

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