## Task A-4.2.

A rubber ball was thrown from a height of 200 meters. Each time it bounces off the surface, it reaches $4 / 5$ of the previous height: after the first bounce, it rises to 160 meters, after the second bounce to 128 meters, etc. What is the total distance the ball travels before it comes to a stop?
## Solution.

The ball first travels 200 meters falling, and then for each bounce, it travels an equal distance upwards and downwards.

Given that

$$
\begin{aligned}
d & =200+2 \cdot 200 \cdot \frac{4}{5}+2 \cdot 200 \cdot\left(\frac{4}{5}\right)^{2}+\cdots \\
& =200+2 \cdot 200 \cdot\left(\frac{4}{5}+\frac{4^{2}}{5^{2}}+\frac{4^{3}}{5^{3}}+\cdots\right) \\
& =200+2 \cdot 200 \cdot\left(\frac{1}{1-\frac{4}{5}}-1\right) \\
& =200+2 \cdot 200 \cdot 4=1800
\end{aligned}
$$

we conclude that the ball travels a total distance of 1800 meters.

Note: If the student multiplies all addends except the first one by 2, or forgets to multiply any factor by 2 (solving the problem as if the ball travels in only one direction), the solution should be graded with 4 points.