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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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    $\sum_{n=0}^\infty h=\infty$; you want to switch the summation and the (-h+) in your last formula.2011-04-30

5 Answers 5

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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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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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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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    It's not rocket surgery either. ;)2011-04-30
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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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    @Wil Alas, this is not the most interesting example for advancing such points. For some better examples see some of my other posts exploiting [innate symmetries.](http://math.stackexchange.com/search?q=user%3A242+innate)2011-04-30