A ball falls from a height of 5,m and strikes | vedclass

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(B) Let the velocity of the ball just before collision be $u_1$ and the velocity of the lift be $u_2$.$1$. Calculate the velocity of the ball before c

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$12\,m/s$ upwards

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(B) Let the velocity of the ball just before collision be $u_1$ and the velocity of the lift be $u_2$.$1$. Calculate the velocity of the ball before collision:Using $v^2 - u^2 = 2gh$,where $u=0$,$g=10\,m/s^2$,and $h=5\,m$:$u_1 = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 5} = 10\,m/s$ (downwards).$2$. Define the coordinate system:Let upwards be positive $(+)$ and downwards be negative $(-)$.So,$u_1 = -10\,m/s$ and $u_2 = +1\,m/s$.$3$. Collision analysis:Since the lift is much heavier than the ball $(m_{lift} \gg m_{ball})$,the lift's velocity remains unchanged after the elastic collision $(e=1)$.Using the relative velocity formula for elastic collision: $v_1 - v_2 = -e(u_1 - u_2)$.Since $v_2 = u_2 = 1\,m/s$:$v_1 - 1 = -1(-10 - 1)$$v_1 - 1 = -1(-11)$$v_1 - 1 = 11$$v_1 = 12\,m/s$.Since the result is positive,the ball rebounds upwards with a velocity of $12\,m/s$.

$A$ ball falls from a height of $5\,m$ and strikes the roof of a lift. If at the time of collision,the lift is moving in the upward direction with a velocity of $1\,m/s$,then the velocity with which the ball rebounds after collision will be (assume $e = 1$):

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