2. Let $h$ be the depth of the well, $g$ be the acceleration due to gravity, and $\tau$ be the interval during which the lid is open. The data is chosen such that $h /\left(g \tau^{2}\right)=1$.

Note that the ball cannot take too long to rise from the well. The maximum time for the ball to rise is $\sqrt{2 h / g}=\tau \sqrt{2}$. Therefore, the ball can only escape during the first opening of the lid after the ball is launched. The speed should not be too high so that the lid can open by the time the ball reaches it ($\tau / 2$). Let's choose the $y$-axis directed upwards with zero at the level of the lid:

$$
y(\tau / 2)=-h+V_{\max } \frac{\tau}{2}-\frac{g \tau^{2}}{8}=0
$$

It is clear that the higher the speed, the later the ball will return. When $V_{\max }=g \tau\left(\frac{2 h}{g \tau^{2}}+\frac{1}{4}\right)$, it will fall on the lid at time $t=4 \tau$. Note that at this moment, the lid is closed. Thus, for different initial speeds, the ball can fall on the lid at times $T \in\left(\frac{3}{2} \tau ; \frac{5}{2} \tau\right) \cup\left(\frac{7}{2} \tau ; 4 \tau\right)$.

We find the corresponding values $\frac{V}{g \tau} \in\left(\frac{2}{3} \frac{h}{g \tau^{2}}+\frac{3}{4}, \frac{2}{5} \frac{h}{g \tau^{2}}+\frac{5}{4}\right) \cup\left(\frac{2}{7} \frac{h}{g \tau^{2}}+\frac{7}{4}, 2 \frac{h}{g \tau^{2}}+\frac{1}{4}\right)$.

Since $\frac{h}{g \tau^{2}}=1$, we get the answer: $\frac{V}{g \tau} \in\left(\frac{17}{12}, \frac{33}{20}\right) \cup\left(\frac{57}{28}, \frac{9}{4}\right)$.
2. The cover of a vertical shaft 160 m deep periodically opens and closes instantly, so that the shaft is open for 4 seconds and closed for 4 seconds. From the bottom of the shaft, a pneumatic gun fires a bullet vertically upwards with an initial velocity $V$, exactly 2 seconds before the next opening of the cover. For what values of the initial velocity $V$ will the bullet freely exit the shaft and fall back onto the cover of the shaft? Assume the acceleration due to gravity is $10 \mathrm{~m} / \mathrm{c}^{2}$.