A ball is dropped from a height h_0. It under | vedclass

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Question

(A) Let $v_0$ be the velocity of the ball just before the first collision and $u_1$ be the velocity just after the first collision. Then $u_1 = e v_0$

Vedclass · Accepted Answer

$e^n = \sqrt{\frac{h_n}{h_0}}$

Vedclass · Answer

(A) Let $v_0$ be the velocity of the ball just before the first collision and $u_1$ be the velocity just after the first collision. Then $u_1 = e v_0$.After the second collision,$u_2 = e u_1 = e^2 v_0$.Continuing this process,after $n$ collisions,the velocity $u_n$ is given by $u_n = e^n v_0$.We know that the velocity of a body falling from height $h$ is $v = \sqrt{2gh}$.Thus,$u_n = \sqrt{2gh_n}$ and $v_0 = \sqrt{2gh_0}$.Substituting these into the equation $u_n = e^n v_0$,we get $\sqrt{2gh_n} = e^n \sqrt{2gh_0}$.Dividing both sides by $\sqrt{2gh_0}$,we obtain $e^n = \sqrt{\frac{h_n}{h_0}}$.

$A$ ball is dropped from a height $h_0$. It undergoes $n$ collisions with the ground. If the velocity of the ball after $n$ rebounds is $u_n$ and it reaches a height $h_n$,which formula gives the coefficient of restitution $e$?

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