5. A small ball was released without initial speed from a height of $h=20 \text{ m}$. The impact with the horizontal surface of the Earth is perfectly elastic. Determine at what moment of time after the start of the fall the average path speed of the ball will be equal to its instantaneous speed. The acceleration due to gravity $g=10 \text{ m} / \text{s}^{2}$. (15 points)
Answer: $2.83 s$

Solution. Obviously, while the ball is flying down, the instantaneous speed will be greater than the average path speed. The moment of time when the ball hits the ground: $t_{1}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \cdot 20}{10}}=2 s$ (2 points). The corresponding speed: $v_{1}=g t_{1}=10 \cdot 2=20 m / s$ (2 points). Equations of motion after a perfectly elastic collision: $x=v_{1}\left(t-t_{1}\right)-\frac{g\left(t-t_{1}\right)^{2}}{2}$ (2 points), $v=v_{1}-g\left(t-t_{1}\right)$ (2 points). According to the condition: $v_{cp}=v$, that is $\frac{h+x}{t}=v \quad(2$ points $), \quad \frac{h+v_{1}\left(t-t_{1}\right)-\frac{g\left(t-t_{1}\right)^{2}}{2}}{t}=v_{1}-g\left(t-t_{1}\right) \quad(2$ points $). \quad$ Solving this equation, we get $t=\sqrt{8} \approx 2.83 s$ (3 points).