# Shorts: Vertical Travel Distance of Bouncing Objects

If a ball is dropped from a height of 10 meters, and on each bounce it reaches a maximal height of 0.75 times the previous height, then what is the total distance traveled? If we use $h_i$ as the maximal height reached on bounce $i$, $h_0$ as the initial height, and $latex b$ as the factor of the maximal height achieved relative to the previous bounce, we have $h_i = b h_{i-1}$.

Then, the total distance traveled is:

\begin{aligned} d &= h_0 + 2 h_1 + 2 h_2 + 2 h_3 + ... \\ &= h_0 + 2 b h_0 + 2 b h_1 + 2 b h_2 + ... \\ &= h_0 + 2 b h_0 + 2 b b h_0 + 2 b b b h_0 + ... \\ &= h_0 + 2 b h_0 + 2 b^2 h_0 + 2 b^3 h_0 + ... \\ &= -h_0 + 2 (h_0 + b h_0 + b^2 h_0 + b^3 h_0 + ...) \\ &= -h_0 + 2 h_0 (1 + b + b^2 + b^3 + ...) \\ &= -h_0 + 2 h_0 \sum_i \left(b^i\right) \end{aligned}

Using $\sum_n \left(x^n\right) = \frac{1}{1-x}$ if $0 \le x < 1$, we get:

$$d = -h_0 + 2 h_0 \frac{1}{1-b}$$

Now we plug our values of $h_0 = 10 \text{ m}$ and $b = 0.75$ into this equation to find $d = -10 + 20 \cdot \frac{1}{0.25} = 70$ meters. So, even though our ball will bounce on for eternity, it will travel exactly 70 meters!

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