A ball is dropped on the floor from a height of 10 m . It rebounds to a height of 2.5 m . If the ball is in contact with the floor for 0.01 sec , the average acceleration during contact is
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Velocity at the time of striking the floor,
\(u = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = 14 m/s\)
Velocity with which it rebounds.
\(v = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 2.5} = 7 \, m/s\)
Change in velocity \(.w = 7 \cdot (-14) = -98 m/s\)
Acceleration \(-\frac{\partial V}{\partial t} - \frac{21}{0.01} = -2100 \quad m/s^{2}\) (upwards)
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