3. The lid of a vertical well 10 m deep periodically opens and closes instantaneously so that the well is in an open state for one second and in a closed state for one second. A stone is thrown vertically upward from the bottom of the well with an initial velocity $V$ exactly 0.5 seconds before the next opening of the lid. For what values of the initial velocity $V$ will the stone freely exit the well and fall back onto the lid of the well? The acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
3. 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 of $\frac{V}{g \tau} \in\left(\frac{2}{3} \frac{h}{g \tau^{2}}+\frac{3}{4}, \frac{2}{5} \frac{h}{g \tau^{2}}+\frac{5}{4}\right) \cup\left(\frac{2}{7} \frac{h}{g \tau^{2}}+\frac{7}{4}, 2 \frac{h}{g \tau^{2}}+\frac{1}{4}\right)$.

Since $\frac{h}{g \tau^{2}}=1$, we get the answer: $\frac{V}{g \tau} \in\left(\frac{17}{12}, \frac{33}{20}\right) \cup\left(\frac{57}{28}, \frac{9}{4}\right)$.

Substituting the numbers, we get the answer:

$V \in\left(\frac{85}{6}, \frac{33}{2}\right) \cup\left(\frac{285}{14}, \frac{45}{2}\right)$