The problem describes the motion of a stone in free fall under constant gravitational acceleration, $g$. Since the stone is dropped, its initial velocity is $v_0 = 0$. The time it takes for the sound to travel back is neglected, so the total time measured, $t$, is equal to the time of the stone's fall.

[Q1] To find the depth of the well, $h$, we use the kinematic equation for displacement under constant acceleration:

$$h = v_0 t + \frac{1}{2} g t^2$$

Given that the stone starts from rest, $v_0 = 0$, the equation simplifies to:

$$h = \frac{1}{2} g t^2$$

We use the standard value for the acceleration due to gravity, $g = 9.8 \text{ m/s}^2$, and the given time, $t = 2.5 \text{ s}$.

Substituting the values into the derived expression:

$$h = \frac{1}{2} (9.8 \text{ m/s}^2) (2.5 \text{ s})^2$$ $$h = (4.9 \text{ m/s}^2) (6.25 \text{ s}^2)$$ $$h = 30.625 \text{ m}$$