91Ó°ÊÓ

The ball is launched from a distance of one yard $\left(h\right)$.

It is given that $n=10$, where n is the number of rebounds.

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Consider a ball is thrown from a height of one yard and it bounces up two-thirds of its original distance. The maximum height (in yards) is calculated after the $n^{th}$bounce.

The height, $n^{th}$rebound $h_n=\left(h\right)×{\left(\frac{2}{3}\right)}^n$

The maximum height after the tenth bounce is,

$$\begin{gathered} h_{10}=\left(1\right)×{\left(\frac{2}{3}\right)}^{10} \\ =0.01734yd \end{gathered}$$

The ball will come in contact with the ground for the tenth time at the end of the ninth rebound. As a result, after the tenth contact with the ground, the distance travelled by the ball is calculated as shown below.

role="math" localid="1654927698887" $$\begin{gathered} D_{n-13}=h+2\left[2\left(1-{\left(\frac{2}{3}\right)}^{\left(n-1\right)}\right)\right] \\ {D}_{n-10}=1+2\left[2\left(1-{\left(\frac{2}{3}\right)}^{\left(10-1\right)}\right)\right] \\ {\mathit{D}}_{\mathbf{9}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{896} \end{gathered}$$

Also, the entire distance travelled by the ball after the tenth bounce is as shown below.

$$\begin{gathered} D_n=h+2\left[2\left(1-{\left(\frac{2}{3}\right)}^{\left(n\right)}\right)\right] \\ {D}_{10}=1+2\left[2\left(1-{\left(\frac{2}{3}\right)}^{\left(10\right)}\right)\right] \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{893}\mathbf{}\mathit{y}\mathit{d} \end{gathered}$$