# Distance traveled by a bouncing ball with exponentially diminishing rebounds

This is kind of an odd question, but can somebody please tell me that I am crazy with the following question, I did the math, and what I am told to prove is simply wrong:

Question: Show that a ball dropped from height of h feet and bounces in such a way that each bounce is $\frac34$ of the height of the bounce before travels a total distance of 7 h feet.

My Work: $$\sum_{n=0}^{\infty} h \left(\frac34\right)^n = 4h$$

Obviously 4 h does not equal 7 h . What does the community get?

I know that my calculations are correct, see Wolfram Alpha and it confirms my calculations, that only leaves my formula, or the teacher being incorrect...

Edit: Thanks everyone for pointing out my flaw, it should be something like: $$\sum_{n=0}^{\infty} -h + 2h \left(\frac34\right)^n = 7h$$

Thanks in advance for any help!

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sigma-algebra is something different than using $\sum$... – Aryabhata Apr 30 '11 at 3:27
Thanks, again, I got the whole n00b thing going on ;) – Nitroware Apr 30 '11 at 3:31
$\sum_{n=0}^\infty h=\infty$; you want to switch the summation and the (-h+) in your last formula. – wildildildlife Apr 30 '11 at 17:29

Note that when the ball bounces it goes both up and down. So from the second term onwards, you need to count each term twice. Therefore the answer is $2 \cdot 4h - h = 7h$ ($h$ is the first term, which is only counted once).

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BINGO! Thanks Yuval and Brandon, this is really an awesome site; how did they manage to get so many rocket scientists on here? – Nitroware Apr 30 '11 at 3:36
@Nitroware: I don't think this is rocket science... ;-) – Fabian Apr 30 '11 at 6:37
It's not rocket surgery either. ;) – J. M. Apr 30 '11 at 10:58

Hint: You are forgetting to count the distance traveled on the way up from the bounce. If the initial drop is from height $h$, the ball bounces up $\frac{3h}{4}$ and then drops from the same distance, and so on.

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Your computation does not give the total distance traveled, it only gives the distance it traveled downward.

The ball first falls $h$. Then it rises $\frac{3}{4}h$, and falls $\frac{3}{4}h$ again; then it rises $(\frac{3}{4})^2h$, and falls that much again. Etc.

So the total distance traveled by the ball is $$h + 2h\left(\frac{3}{4}\right) + 2h\left(\frac{3}{4}\right)^2 + \cdots = h + \sum_{n=0}^{\infty} \frac{3h}{2}\left(\frac{3}{4}\right)^{n-1}.$$

This gives, by the usual formula, a total distance of: $$h + \frac{\quad\frac{3h}{2}\quad}{1 - \frac{3}{4}} = h + \frac{\quad\frac{3h}{2}\quad}{\frac{1}{4}} = h + \frac{12h}{2} = 7h.$$

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To get symmetry - a complete sawtooth pattern - suppose the ball is first thrown up from ground to height $h$. The total distance $x$ is that of the first tooth $= 2h$ plus the remaining sawtooth $= 3/4\ x$. Thus $x =\: 2h + 3/4\ x,\$ so $x = 8h\:.\$ Subtracting the intitial throw up leaves $7h.$

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@Bill: at first sight this looks very elegant because it seems to avoid the use of geometric series. However, I wonder how to make the observation 'the remaining sawtooth = $3/4 x$' rigourous without resorting to geometric series after all. – wildildildlife Apr 30 '11 at 17:31
@wil Indeed, there are various ways to make the argument rogorous, depending on the OP's background. But my goal was to encourage students to think about the innate symmetry, here the shift that arises by mutiplication by $3/4\:.\:$ Shift symmetries are ubiquitous so, it lends insight to bring them to the fore when they play fundamental roles. – Bill Dubuque Apr 30 '11 at 17:49
Ok, so then you're actually rederiving the geometric series using the shift argument: $S=\sum_k r^k\Rightarrow S-rS= 1\Rightarrow S=\frac{1}{1-r}$. – wildildildlife Apr 30 '11 at 17:56
@Wil That's one way to view it. But rather than teaching students to be able to rotely sum geometric series, I prefer to teach them the essence of the matter, here the innate shift symmetry. Exploiting innate symmetries is a fundamental mathematical problem-solving technique. So I alway try to encourage viewpoints that help to develop intuition for such. – Bill Dubuque Apr 30 '11 at 18:51
@Bill: I appreciate your didactical ideas! – wildildildlife Apr 30 '11 at 19:39

Before jumping to a formula, let us calculate a little. The distance travelled until the first contact with the ground is $h$.

The distance travelled between the first contact and the second is $(h)(2)(3/4)$ (up and then down). The distance travelled from second contact to third is $(h)(2)(3/4)^2$, and so on.

So the total distance travelled is $$h+(h)(2)\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^n$$

Finally, sum the infinite series. That sum is $3$, giving a total of $7h$.

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