MHB How Far Does a Bouncing Ball Travel Before Stopping?

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A ball dropped from a height of 270 m rebounds to 10% of its previous height with each bounce. The total vertical distance traveled by the ball before coming to rest is calculated using a geometric series. The formula derived shows that the total distance is 330 m. This includes the initial drop and all subsequent rebounds. The calculations confirm that the answer of 330 m is correct.
Starkiller2301
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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
 
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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.}

 
soroban said:
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!
 
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