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(A) The velocity of the ball just before hitting the floor is $v_i = \sqrt{2gh_i} = \sqrt{2 	imes 10 	imes 9.8} = \sqrt{196} = 14\,m/s$ (downwards).

Vedclass · Accepted Answer

$120$

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(A) The velocity of the ball just before hitting the floor is $v_i = \sqrt{2gh_i} = \sqrt{2 	imes 10 	imes 9.8} = \sqrt{196} = 14\,m/s$ (downwards).Taking the downward direction as negative,$v_i = -14\,m/s$.The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_f} = \sqrt{2 	imes 10 	imes 5} = \sqrt{100} = 10\,m/s$ (upwards).Taking the upward direction as positive,$v_f = +10\,m/s$.The change in velocity is $\Delta v = v_f - v_i = 10 - (-14) = 24\,m/s$.The average acceleration is $a_{	ext{avg}} = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120\,m/s^2$.

Column-$I$	Column-$II$
$(1)$ Change in momentum	$(a)$ Force
$(2)$ Rate of change of momentum	$(b)$ Impulse

$A$ tennis ball is dropped onto the floor from a height of $9.8\,m$. It rebounds to a height of $5.0\,m$. The ball comes in contact with the floor for $0.2\,s$. The average acceleration during contact is $..........\,m/s^2$. [Given $g = 10\,m/s^2$]

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