A particle falls from a height h upon a fixed | vedclass

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Question

(A) The particle falls from height $h$. The distance covered in the first fall is $h$.After the first impact,the velocity of rebound is $v_1 = e \sqrt

Vedclass · Accepted Answer

$h\left(\frac{1+e^2}{1-e^2}\right)$

Vedclass · Answer

(A) The particle falls from height $h$. The distance covered in the first fall is $h$.After the first impact,the velocity of rebound is $v_1 = e \sqrt{2gh}$. The height reached after the first rebound is $h_1 = \frac{v_1^2}{2g} = e^2 h$.The particle travels this distance $h_1$ upwards and $h_1$ downwards,so the distance covered is $2h_1 = 2e^2 h$.After the second impact,the height reached is $h_2 = e^2 h_1 = e^4 h$. The distance covered is $2h_2 = 2e^4 h$.This continues as a geometric progression: $h + 2e^2 h + 2e^4 h + 2e^6 h + \dots$Total distance $D = h + 2e^2 h (1 + e^2 + e^4 + \dots)$.Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$,where $a = 1$ and $r = e^2$:$D = h + 2e^2 h \left(\frac{1}{1-e^2}\right) = h \left(1 + \frac{2e^2}{1-e^2}\right) = h \left(\frac{1-e^2+2e^2}{1-e^2}\right) = h \left(\frac{1+e^2}{1-e^2}\right)$.

$A$ particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before rebounding has stopped is

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