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

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Question

(B) When the ball is dropped from height $h$,it hits the floor for the first time after falling a distance $h$.After the first collision,the ball rebo

Vedclass · Accepted Answer

$h(1+2e^2)$

Vedclass · Answer

(B) When the ball is dropped from height $h$,it hits the floor for the first time after falling a distance $h$.After the first collision,the ball rebounds to a height $h_1 = e^2 h$.The ball then travels upwards to height $h_1$ and then falls back down to the floor to hit it for the second time.Therefore,the total distance covered by the ball just before the second hit is the sum of the initial fall distance,the rebound height,and the distance fallen from the rebound height.Total distance $D = h + h_1 + h_1 = h + 2h_1$.Substituting $h_1 = e^2 h$ into the equation,we get $D = h + 2(e^2 h) = h(1 + 2e^2)$.

$A$ ball is dropped from a height $h$ on a floor of coefficient of restitution $e$. The total distance covered by the ball just before the second hit is

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