A ball is thrown with a velocity of 6, m/s ve | vedclass

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(D) Let $u = 6\, m/s$ be the initial downward velocity and $H = 3.2\, m$ be the height.First,calculate the velocity $v$ of the ball just before hittin

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$0.8$

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(D) Let $u = 6\, m/s$ be the initial downward velocity and $H = 3.2\, m$ be the height.First,calculate the velocity $v$ of the ball just before hitting the floor using the equation $v^2 = u^2 + 2gH$:$v^2 = 6^2 + 2 	imes 10 	imes 3.2 = 36 + 64 = 100$$v = 10\, m/s$.Let $v'$ be the velocity of the ball just after the rebound. Since the ball rebounds to the same height $H$,the velocity required to reach height $H$ is $v' = \sqrt{2gH} = \sqrt{2 	imes 10 	imes 3.2} = \sqrt{64} = 8\, m/s$.The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach:$e = \frac{v'}{v} = \frac{8}{10} = 0.8$.

$A$ ball is thrown with a velocity of $6\, m/s$ vertically downwards from a height $H = 3.2\, m$ above a horizontal floor. If it rebounds back to the same height,then the coefficient of restitution $e$ is $[g = 10\, m/s^2]$.

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