A ball falls from a height h upon a fixed hor | vedclass

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Question

(A) The ball falls from height $h$ and hits the floor. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.After the first impact, the vel

Vedclass · Accepted Answer

$h \frac{1+e^2}{1-e^2}$

Vedclass · Answer

(A) The ball falls from height $h$ and hits the floor. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.After the first impact, the velocity becomes $v_1 = e v_0$. The ball rises to a height $h_1 = \frac{v_1^2}{2g} = e^2 h$.It then falls from $h_1$ and hits the floor again. The distance covered in the first bounce (up and down) is $2h_1 = 2e^2 h$.After the second impact, it rises to $h_2 = e^2 h_1 = e^4 h$. The distance covered in the second bounce is $2h_2 = 2e^4 h$.The total distance $D$ is the sum of the initial fall and all subsequent bounces:$D = h + 2h_1 + 2h_2 + 2h_3 + ...$$D = h + 2e^2 h + 2e^4 h + 2e^6 h + ...$$D = h + 2e^2 h (1 + e^2 + e^4 + ...)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:$D = h + 2e^2 h \left( \frac{1}{1-e^2} \right)$$D = h \left( 1 + \frac{2e^2}{1-e^2} \right) = h \left( \frac{1-e^2+2e^2}{1-e^2} \right) = h \left( \frac{1+e^2}{1-e^2} \right)$.

$A$ ball falls from a height $h$ upon a fixed horizontal floor. The coefficient of restitution for the collision between the ball and the floor is $e$. The total distance covered by the ball before coming to rest is [neglect the air resistance].

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