1. **Problem Statement:** Determine how many full bounces a ball makes within 20 seconds, given it bounces to 80% of its previous height each time and the initial drop height is 10 m. 2. **Formula for time of each bounce:** The time for a bounce from height $h$ is $$ t = 2\sqrt{\frac{2h}{g}} $$ where $g = 10$ m/s$^2$. 3. **Initial bounce time:** For $h_0 = 10$ m, $$ t_1 = 2\sqrt{\frac{2 \times 10}{10}} = 2\sqrt{2} \approx 2.8284 \text{ s} $$ 4. **Subsequent bounce heights and times:** Each bounce height is $$ h_n = 0.8^n \times 10 $$ Time for $n$-th bounce: $$ t_n = 2\sqrt{2 \times 0.8^n} = 2\sqrt{2} \times (0.8^{n/2}) $$ 5. **Total time for $n$ bounces:** Sum of geometric series $$ T_n = 2\sqrt{2} \sum_{k=0}^{n-1} (0.8^{k/2}) = 2\sqrt{2} \cdot \frac{1 - (0.8^{n/2})}{1 - \sqrt{0.8}} $$ 6. **Find max $n$ such that $T_n \leq 20$:** $$ 2\sqrt{2} \cdot \frac{1 - (0.8^{n/2})}{1 - \sqrt{0.8}} \leq 20 $$ Divide both sides: $$ \frac{1 - (0.8^{n/2})}{1 - \sqrt{0.8}} \leq 5\sqrt{2} $$ Rearranged: $$ 0.8^{n/2} \geq 1 - 5\sqrt{2}(1 - \sqrt{0.8}) $$ Calculate constants: $$ \sqrt{0.8} \approx 0.8944, \quad 1 - \sqrt{0.8} \approx 0.1056 $$ $$ 5\sqrt{2}(1 - \sqrt{0.8}) \approx 5 \times 1.4142 \times 0.1056 \approx 0.746 $$ So: $$ 0.8^{n/2} \geq 0.254 $$ Take logarithms: $$ \frac{n}{2} \log(0.8) \leq \log(0.254) $$ $$ n \leq \frac{2 \log(0.254)}{\log(0.8)} $$ Using $$ \log(0.254) \approx -0.5945, \quad \log(0.8) \approx -0.09691 $$ $$ n \leq \frac{2 \times -0.5945}{-0.09691} \approx 12.25 $$ 7. **Final answer:** The maximum full bounces within 20 seconds is $$ \boxed{12} $$