A ball after falling from a height of 10, m s | vedclass

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(C) The speed of the ball just before striking the lift is given by $v = \sqrt{2gh}$.Taking $g = 9.8\, m/s^2$ and $h = 10\, m$,we get $v = \sqrt{2 	i

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The speed of the ball just before striking the lift is given by $v = \sqrt{2gh}$.Taking $g = 9.8\, m/s^2$ and $h = 10\, m$,we get $v = \sqrt{2 	imes 9.8 	imes 10} = \sqrt{196} = 14\, m/s$.Let the downward direction be positive. The velocity of the ball before collision is $u_1 = 14\, m/s$ and the velocity of the lift is $u_2 = 1\, m/s$.Since the mass of the lift is much greater than the mass of the ball $(M_{lift} \gg M_{ball})$,the collision is effectively an elastic collision with a stationary wall moving at velocity $u_2$.The velocity of the ball after collision $v_1$ is given by the formula $v_1 = 2u_2 - u_1$ (relative to the ground).Substituting the values: $v_1 = 2(1) - 14 = 2 - 14 = -12\, m/s$.The negative sign indicates that the ball is moving upwards with a speed of $12\, m/s$.

$A$ ball after falling from a height of $10\, m$ strikes the roof of a lift which is descending down with a velocity of $1\, m/s$. The recoil velocity of the ball will be .............. $m/s$.

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