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,
\(u = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 10} = 14 \, m/s\)
Velocity with which it rebounds.
\(v = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 2.5} = 7 \, m/s\)
Change in velocity \(\Delta v = 7 - (-14) = 21\,m/s\)
Acceleration \(\frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 \ \mathrm{m/s^{2}}\) (upwards)
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