A ball is let fall from a height h_0. It make | vedclass

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(A) The velocity of the ball just before the first collision is $v_0 = \sqrt{2gh_0}$.After the first collision,the velocity is $v_1 = ev_0$.After the

Vedclass · Accepted Answer

$e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}$

Vedclass · Answer

(A) The velocity of the ball just before the first collision is $v_0 = \sqrt{2gh_0}$.After the first collision,the velocity is $v_1 = ev_0$.After the second collision,the velocity is $v_2 = ev_1 = e^2v_0$.Following this pattern,after $n$ collisions,the velocity of the ball is $v_n = e^n v_0$.The height $h_n$ reached after the $n$th collision is given by $h_n = \frac{v_n^2}{2g}$.Substituting $v_n = e^n v_0$,we get $h_n = \frac{(e^n v_0)^2}{2g} = e^{2n} \frac{v_0^2}{2g}$.Since $h_0 = \frac{v_0^2}{2g}$,we have $h_n = e^{2n} h_0$.Therefore,$e^{2n} = \frac{h_n}{h_0}$,which implies $e = \left[\frac{h_n}{h_0}\right]^{1/2n}$.

$A$ ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity $v_n$ and the ball rises to a height $h_n$. Then the coefficient of restitution is given by:

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