Q74 / 90 · KVPY 2010 (SB/SX) · Physics
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 traveled by the ball up to the $10^{\text {th }}$ impact ?
- A$2 \mathrm{~h} \frac{1-\mathrm{r}^{10}}{1-\mathrm{r}}$
- B$\mathrm{h} \frac{1-\mathrm{r}^{20}}{1-\mathrm{r}^{2}}$
- C$2 \mathrm{~h} \frac{1-\mathrm{r}^{22}}{1-\mathrm{r}^{2}}-\mathrm{h}$
- Correct answerD$2 \mathrm{~h} \frac{1-\mathrm{r}^{20}}{1-\mathrm{r}^{2}}-\mathrm{h}$
Solution
$\quad$ Total distance $=\left(\frac{\mathrm{v}_{0}^{2}}{g}+\mathrm{r}^{2} \frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}+\mathrm{r}^{4} \frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}+\ldots\right.$ upto $10^{\text {th }}$ terms $)-\mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}\left(1+\mathrm{r}^{2}+\mathrm{r}^{4}+\ldots .+10^{\text {th }}\right.$ term $)-\mathrm{h}$
also $v_{0}=\sqrt{2 g h}$
$$
\begin{array}{l}
\therefore \text { Total distance }=2 \mathrm{~h}\left(\frac{1-\left(\mathrm{r}^{2}\right)^{10}}{1-\mathrm{r}^{2}}\right)-\mathrm{h} \\
\text { or } \text { total distance }=\frac{2 \mathrm{~h}\left(1-\mathrm{r}^{20}\right)}{\left(1-\mathrm{r}^{2}\right)}-\mathrm{h}
\end{array}
$$
Hence, the correct answer is option: 'D'.