In the bouncing ball example above, find the height of the tenth rebound, and the distance traveled by the ball after it touches the ground the tenth time. Compare this distance with the total distance traveled.

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 $h_n=\left(h\right)\times {\left(\frac{2}{3}\right)}^n$

The maximum height after the tenth bounce is,

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

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.

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