How Far Does a Bouncing Ball Travel Before Stopping?

[Starkiller2301]

Hi everybody, can you please help me with this question and with the working out?

A ball was dropped from a height of 270 m. On each rebound, it rose to 10% of the previous height. Find the total vertical distance traveled by the ball before coming to rest.
Thanks,
Starkiller2301

 

[soroban]

Hello, Starkiller2301!

A ball was dropped from a height of 270 m.
On each rebound, it rose to 10% of the previous height.
Find the total vertical distance traveled by the ball before coming to rest.

Let $$x$$ = original height.

First, the ball falls $$x$$ meters.

It bounces up $$\tfrac{x}{10}$$ m, and falls $$\tfrac{x}{10}$$ m.
It bounces up $$\tfrac{x}{10^2}$$ m, and falls $$\tfrac{x}{10^2}$$ m.
It bounces up $$\tfrac{x}{10^3}\,m$$, and falls $$\tfrac{x}{10^3}\,m.$$
And so on.Total distance:

$$\quad d \;=\;x + 2(\tfrac{x}{10}) + 2(\tfrac{x}{10^2}) + 2(\tfrac{x}{10^3}) \cdots$$

$$\quad d \;=\;x\left[1 + \tfrac{2}{10}\underbrace{\left(1 + \tfrac{1}{10} + \tfrac{1}{10^2} +\tfrac{1}{10^3} + \cdots \right)}_{\text{geometric series}} \right]$$

The geometric series has sum $$\frac{1}{1-\frac{1}{10}} \:=\: \frac{1}{\frac{9}{10}} \:=\:\frac{10}{9}$$

$$d\;=\;x\left[1 + \tfrac{2}{10}\left(\tfrac{10}{9}\right)\right] \;=\;\tfrac{11}{9}x$$

Therefore: $$\;d \;=\;\tfrac{11}{9}(270) \;=\;330\text{ m.}$$

 

[Starkiller2301]

Hello, Starkiller2301!


Let $$x$$ = original height.

First, the ball falls $$x$$ meters.

It bounces up $$\tfrac{x}{10}$$ m, and falls $$\tfrac{x}{10}$$ m.
It bounces up $$\tfrac{x}{10^2}$$ m, and falls $$\tfrac{x}{10^2}$$ m.
It bounces up $$\tfrac{x}{10^3}\,m$$, and falls $$\tfrac{x}{10^3}\,m.$$
And so on.Total distance:

$$\quad d \;=\;x + 2(\tfrac{x}{10}) + 2(\tfrac{x}{10^2}) + 2(\tfrac{x}{10^3}) \cdots$$

$$\quad d \;=\;x\left[1 + \tfrac{2}{10}\underbrace{\left(1 + \tfrac{1}{10} + \tfrac{1}{10^2} +\tfrac{1}{10^3} + \cdots \right)}_{\text{geometric series}} \right]$$

The geometric series has sum $$\frac{1}{1-\frac{1}{10}} \:=\: \frac{1}{\frac{9}{10}} \:=\:\frac{10}{9}$$

$$d\;=\;x\left[1 + \tfrac{2}{10}\left(\tfrac{10}{9}\right)\right] \;=\;\tfrac{11}{9}x$$

Therefore: $$\;d \;=\;\tfrac{11}{9}(270) \;=\;330\text{ m.}$$

Thanks so much! I had the answer sheet but I didn't know how to get the answer. 330m was correct!