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(D) Let $H_0 = 6.25 \ m$ be the initial height and $H_2 = 81 \ cm = 0.81 \ m$ be the height reached after the second bounce.For a ball bouncing on a h

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$0.6$

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(D) Let $H_0 = 6.25 \ m$ be the initial height and $H_2 = 81 \ cm = 0.81 \ m$ be the height reached after the second bounce.For a ball bouncing on a hard surface,the height reached after $n$ bounces is given by $H_n = H_0 \cdot e^{2n}$,where $e$ is the coefficient of restitution.Given $n = 2$,$H_0 = 6.25 \ m$,and $H_2 = 0.81 \ m$.Substituting the values:$0.81 = 6.25 \cdot e^{2 	imes 2}$$0.81 = 6.25 \cdot e^4$$e^4 = \frac{0.81}{6.25} = \frac{81}{625}$Taking the square root on both sides:$e^2 = \sqrt{\frac{81}{625}} = \frac{9}{25} = 0.36$Taking the square root again:$e = \sqrt{0.36} = 0.6$Thus,the coefficient of restitution is $0.6$.

$A$ ball falls freely from rest from a height of $6.25 \ m$ onto a hard horizontal surface. If the ball reaches a height of $81 \ cm$ after the second bounce from the surface,the coefficient of restitution is

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