1. **Problem statement:** A tennis ball is dropped from a ceiling height $h_0=10$ m. Each bounce reaches 80% of the previous height. The time for a full bounce (up and down) from height $h$ is given by $$t = 2\sqrt{\frac{2h}{g}}$$ with $g=10$ m/s$^2$. We want to find how many full bounces occur within 20 seconds. 2. **Understanding the problem:** A full bounce means the ball goes from the ceiling down to the floor and back up to the ceiling height it reached after bouncing. The initial drop is from $h_0=10$ m, then each bounce height is $0.8$ times the previous height. 3. **Time for the first full bounce:** The ball falls from $10$ m and bounces back up to $0.8 \times 10 = 8$ m. The time for the first full bounce is the sum of the time to fall from 10 m and the time to go up and down from 8 m. - Time to fall from 10 m (just down): $$t_{fall} = \sqrt{\frac{2h_0}{g}} = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2}$$ - Time for the first bounce up and down from 8 m: $$t_1 = 2\sqrt{\frac{2 \times 8}{10}} = 2\sqrt{1.6}$$ - Total time for first full bounce: $$T_1 = t_{fall} + t_1 = \sqrt{2} + 2\sqrt{1.6}$$ 4. **Time for subsequent bounces:** For the $n$-th bounce, the height is $$h_n = 0.8^n \times 10$$ and the time for the bounce up and down is $$T_n = 2\sqrt{\frac{2h_n}{g}} = 2\sqrt{\frac{2 \times 10 \times 0.8^n}{10}} = 2\sqrt{2 \times 0.8^n} = 2\sqrt{2} \times (0.8)^{n/2}$$ 5. **Total time after $N$ full bounces:** $$ S_N = t_{fall} + \sum_{n=1}^N T_n = \sqrt{2} + \sum_{n=1}^N 2\sqrt{2} (0.8)^{n/2} = \sqrt{2} + 2\sqrt{2} \sum_{n=1}^N (0.8)^{n/2} $$ 6. **Sum of geometric series:** Let $r = (0.8)^{1/2} = \sqrt{0.8}$, then $$ \sum_{n=1}^N r^n = r \frac{1-r^N}{1-r} $$ So, $$ S_N = \sqrt{2} + 2\sqrt{2} \times r \frac{1-r^N}{1-r} = \sqrt{2} + 2\sqrt{2} \times \sqrt{0.8} \frac{1-(\sqrt{0.8})^N}{1-\sqrt{0.8}} $$ 7. **Find maximum $N$ such that $S_N \leq 20$ seconds:** $$ \sqrt{2} + 2\sqrt{2} \sqrt{0.8} \frac{1-(\sqrt{0.8})^N}{1-\sqrt{0.8}} \leq 20 $$ Rearranged: $$ \frac{1-(\sqrt{0.8})^N}{1-\sqrt{0.8}} \leq \frac{20 - \sqrt{2}}{2\sqrt{2} \sqrt{0.8}} $$ Calculate constants: - $\sqrt{2} \approx 1.414$ - $\sqrt{0.8} \approx 0.8944$ - $1 - \sqrt{0.8} = 0.1056$ - Numerator: $20 - 1.414 = 18.586$ - Denominator: $2 \times 1.414 \times 0.8944 \approx 2.53$ So, $$ \frac{1-(0.8944)^N}{0.1056} \leq \frac{18.586}{2.53} \approx 7.35 $$ Multiply both sides by $0.1056$: $$ 1 - (0.8944)^N \leq 0.776 $$ Thus, $$ (0.8944)^N \geq 1 - 0.776 = 0.224 $$ Take natural logarithm: $$ N \ln(0.8944) \geq \ln(0.224) $$ Since $\ln(0.8944) < 0$, divide inequality by negative number flips inequality: $$ N \leq \frac{\ln(0.224)}{\ln(0.8944)} $$ Calculate: - $\ln(0.224) \approx -1.495$ - $\ln(0.8944) \approx -0.1116$ So, $$ N \leq \frac{-1.495}{-0.1116} \approx 13.4 $$ 8. **Final answer:** The number of full bounces within 20 seconds is the greatest integer less than or equal to 13.4, which is $$\boxed{13}$$