# Thread: Bouncing Ball Series

1. ## Bouncing Ball Series

Having trouble solving a problem regarding the bouncing ball series in terms of the time needed for the ball to stop bouncing.

We are to first find an equation for time (t) in terms of height (h)

I had used the formula h=g(gravity) * (1/2)t^2, gravity being in English units of measurement so approximately 32 ft/sec

That came to t = (1/4) sqrt(h)

We then are given height h = 12 ft and are told to compute how long it will take for the ball to stop bouncing given the ball bounces back up 3/4 of the height of the previous height.

Using h = 12 and r(common ratio) = 3/4, I then used the formula S = a / (1 - r) and came up with 2sqrt(3) seconds, approximately 3.464 seconds.

I didn't feel all to confident in my methods to begin with but based upon a short conversation with classmates, I am nearly certain I have gone about this incorrectly.

Any pointers towards the right way to go about solving this would be much appreciated and any detailed breakdown to accompany it would be fantastic.

2. ## Re: Bouncing Ball Series

surely if every bounce is 0.75 times the previous one, it will never stop bouncing?

3. ## Re: Bouncing Ball Series

Hello, TheNotoriousWMB!

A problem regarding a bouncing ball and the time needed for the ball to stop bouncing.

We are to first find an equation for time (t) in terms of height (h).

I used the formula: $\displaystyle h\:=\:16t^2$

We are given the initial height h = 12 ft
and that the ball bounces up 3/4 of the previous height.

We are told to compute how long it will take for the ball to stop bouncing.

Contrary to our intuition, the ball stops bouncing in a finite length of time.

Consider the total distance the balls travels.

It falls 12 feet.

It bounces up $\displaystyle (\tfrac{3}{4})(12)$ feet.
It falls $\displaystyle (\tfrac{3}{4})(12)$ feet.

It bounces up $\displaystyle (\tfrac{3}{4})^2(12)$ feet.
It falls $\displaystyle (\tfrac{3}{4})^2(12)$ feet.

It bounces up $\displaystyle (\tfrac{3}{4})^3(12)$ feet.
It falls $\displaystyle (\tfrac{3}{4})^3(12)$ feet.

. . . and so on.

The total distance is:

. . $\displaystyle D \;=\;12 + 2(\tfrac{3}{4})(12) + 2(\tfrac{3}{4})^2(12) + 2(\tfrac{3}{4})^3(12) + \hdots$

. . $\displaystyle D \;=\;12 + 2(\tfrac{3}{4})(12)\!\cdot\!\underbrace{\left[1 + \tfrac{3}{4} + (\tfrac{3}{4})^2 + (\tfrac{3}{4})^3 + \hdots \right]}_{\text{geometric series}}$

The geometric series has the sum: .$\displaystyle \frac{1}{1-\frac{3}{4}} \:=\:4$

Therefore: .$\displaystyle D \;=\;12 + 18(4) \;=\;84\text{ feet.}$

Since the ball travels a finite distance, it must take a finite amount of time.

4. ## Re: Bouncing Ball Series

The time to go up and down the first time is 2 times $\displaystyle \sqrt{\frac{h}{16}}=$$\displaystyle \sqrt{\frac{12}{16}}=\frac{\sqrt{3}}{2}$ or $\displaystyle \sqrt{3}$ seconds.

The second time it takes 3/2 seconds. The third time it takes $\displaystyle \frac{3\sqrt{3}}{4}$, and so on, so the nth time takes $\displaystyle \sqrt{3}\left(\frac{\sqrt{3}}{2}\right)^n$ seconds, so the total time is the sum of the series:$\displaystyle S=\frac{a}{1-r}=\frac{\sqrt{3}}{1-\frac{\sqrt{3}}{2}}$ or approximately 12.9 seconds.

Soroban, I think your final step is invalid. It is possible for an object to slow down in such a way that it approaches a finite distance as the time goes to infinity. But that's not what happens in this problem.

- Hollywood

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### bouncing ball problem geometric series

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