A ball hits horizontally at 6.0ms-1. It rebounds horizontally at 4.4 ms-1. The ball is in contact with the wall for 0.04 s. What is the acceleration of the ball .
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Solution: Initial velocity of a ball (u) = 6.0 m/s Final velocity is an opposite direction to the initial velocity (v) = - 4.4 m/s Time taken (t) = 0.04 sec \\ \text{Acceleration of the ball } = \frac{\text{Final velocity (v) - Initial velocity (u)}}{\text{Time taken(t)}}
=\frac{-4.4-6.0}{0.04}=\frac{-10.4}{0.04}=-260m/s^2
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