# Superball total bounce distance

I am asked to explain how to calculate total bounce distance:

A "super" rubber ball is one that is measured to bounce when dropped 70% or higher that the distance from which it is dropped.

You are to take a super rubber ball that bounces 75% of it dropped height and you are to find out the total distance traveled by this ball when dropped from the top of a 100 foot building.

Careful with your calculations in the very beginning of this experiment.

I planned on working out the problem by taking 75% from each bounce util the height was near or equal to 0 then add each bounce together for the total.

Is there an easier way to do this. Thanks for the help.

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If you drop the ball from height $h$, it bounces up ${3\over 4}h$. – ncmathsadist Nov 24 '12 at 17:45

You first drop your ball from height $x_0 = 100$, and it bounces back to height $x_1 = \frac{75}{100} \times x_0$. From there, it will drop again, from height $x_1$ this time, and bounce back again to height $x_2 = \frac{75}{100} \times x_1$. So basically, what you want to compute is $$x_0 + \sum_{i=1}^{\infty} 2 \times x_i = 2 \times \left(\sum_{i=0}^{\infty} x_i\right) - x_0.$$ This is a geometric series.
It's a geometric series starting with the second term (the first term doesn't fit the pattern; just add it in separately): $$a + 0.75(2a) + 0.75^2(2a) + 0.75^3(2a)+\cdots.$$ The sum of a geometric series is well known.
At first, the ball falls down $100$ feet. After the $k$th bounce it ascends to a height of $(0.75)^k \times 100$ feet and falls down the same distance. So its total travelling distance is $$100 \text{ feet} \times \left(1+2 \times \sum_{k=1}^\infty (0.75)^k\right)=100 \text{ feet} \times \left(1+2 \times \frac{0.75}{1-0.75}\right)=700\text{ feet.}$$