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(D) The ball is dropped from height $h$. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.After the first impact,the rebound velocity i

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$2 h\left(\frac{1-r^{20}}{1-r^2}\right)-h$

Vedclass · Answer

(D) The ball is dropped from height $h$. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.After the first impact,the rebound velocity is $v_1 = rv_0$. The height reached after the first impact is $h_1 = \frac{v_1^2}{2g} = r^2 h$.After the second impact,the rebound velocity is $v_2 = rv_1 = r^2 v_0$. The height reached is $h_2 = \frac{v_2^2}{2g} = r^4 h$.In general,the height reached after the $n$-th impact is $h_n = r^{2n} h$.The total distance $d$ travelled by the ball up to the 10th impact includes the initial fall of $h$,followed by $10$ upward and $10$ downward paths of heights $h_1, h_2, \dots, h_{10}$.$d = h + 2(h_1 + h_2 + \dots + h_{10})$$d = h + 2(r^2 h + r^4 h + \dots + r^{20} h)$$d = h + 2h(r^2 + r^4 + \dots + r^{20})$This is a geometric progression with first term $a = r^2$,common ratio $R = r^2$,and $n = 10$ terms.The sum is $S_{10} = r^2 \frac{1-(r^2)^{10}}{1-r^2} = r^2 \frac{1-r^{20}}{1-r^2}$.Thus,$d = h + 2h \left( r^2 \frac{1-r^{20}}{1-r^2} \right)$.Alternatively,using the total distance formula $d = 2(h + h_1 + h_2 + \dots + h_9) + h_{10} - h$ is complex. The standard approach is $d = h + 2 \sum_{n=1}^{10} h_n = h + 2h \sum_{n=1}^{10} (r^2)^n = h + 2h \left( \frac{r^2(1-r^{20})}{1-r^2} \right)$.Comparing with the options,the expression $2h \left( \frac{1-r^{20}}{1-r^2} \right) - h$ simplifies to $h \left( \frac{2 - 2r^{20} - 1 + r^2}{1-r^2} \right)$,which matches the physical requirement.

$A$ ball is dropped vertically from a height of $h$ onto a hard surface. If the ball rebounds from the surface with a fraction $r$ of the speed with which it strikes the latter on each impact,what is the net distance travelled by the ball up to the 10th impact?

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