A ball is dropped on the floor from a height | vedclass

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(B) Velocity at the time of striking the floor $(u_1)$: $u_1 = \sqrt{2gh_1} = \sqrt{2 	imes 9.8 	imes 10} = \sqrt{196} = 14 \, m/s$ (downwards). Tak

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$2100 \, m/s^2$ upwards

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(B) Velocity at the time of striking the floor $(u_1)$: $u_1 = \sqrt{2gh_1} = \sqrt{2 	imes 9.8 	imes 10} = \sqrt{196} = 14 \, m/s$ (downwards). Taking downward direction as negative,$u_1 = -14 \, m/s$. Velocity with which it rebounds $(v_1)$: $v_1 = \sqrt{2gh_2} = \sqrt{2 	imes 9.8 	imes 2.5} = \sqrt{49} = 7 \, m/s$ (upwards). Taking upward direction as positive,$v_1 = +7 \, m/s$. Change in velocity $\Delta v = v_1 - u_1 = 7 - (-14) = 21 \, m/s$. Average acceleration $a = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 \, m/s^2$. Since the change in velocity is positive,the acceleration is directed upwards.

$A$ 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 $0.01 \, s$,the average acceleration during contact is:

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