Puzzle – Leaning Tower of Pisa

April 2, 2010 – 10:55 pm

Let’s say you drop a ball from the Leaning Tower of Pisa, which is 179 ft high, and it bounces back 10% of the dropped height – 17.9 ft. Then on the second bounce it bounces up 10% again – 1.79 ft, and so on for ever. How far will the ball travel?

The first thing to realize is that this is a geometric progression which will yield a finite sum, even though it has an infinite number of terms. This idea of finite sums is the reason Zeno can walk across the room: http://en.wikipedia.org/wiki/Geometric_series#Zeno.27s_paradoxes.

Before answering the question it may be useful to recap geometric progressions from my previous post see the coin-toss puzzle answer.

In this problem, the total distance travelled by the ball will be: 179 + 179 * 2 * (1/10 + 1/100 + 1/1000…)

You may be able to see straight away that the sum of the progression is 1/9, and so the total distance 218 + 7/9.

But let’s do it using the standard approach:

Given that n in the progression will start at 1, we need to get our progression into the form:

<br /> \sum_{n=1}^\infty cr^{n-1} = \frac{c}{1-r}<br />

When we do this, we get the same answer as above:

<br /> distance =\;179\;+\;179\;*\;2\;*\;1/10\;*\;\frac{1}{1-\frac{1}{10}}\;=\;218\;\frac{7}{9}<br />

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