13.333. A ball falls from a height of 2 m 43 cm and, upon hitting the ground, bounces back up, each time reaching $2 / 3$ of the height from which it falls again. After how many bounces will the ball rise to a height of 32 cm?
Solution.

The numbers expressing the height to which the ball rises form a geometric progression with $b_{1}=243 q$ and $q=\frac{2}{3}$. According to the condition $q_{n}=32=b_{1} q^{n-1}$, from which $n=5$.

Answer: after 5 bounces.