Stone in a Well Problem

To find the time $t_{1}$ it takes for the stone to fall to the bottom of the well, we use the kinematic equation for an object in free fall under constant acceleration. Let's define the downward direction as positive. The displacement of the stone is the depth of the well, $h$. The stone is dropped, so its initial velocity is $v_0 = 0$. The acceleration is the acceleration due to gravity, $a=g$.

The kinematic equation for displacement is:

$h = v_{0}t_{1} + \frac{1}{2}at_{1}^2$

Substituting the known values for the stone's fall:

$h = (0) \cdot t_{1} + \frac{1}{2}g t_{1}^2$

$h = \frac{1}{2}gt_{1}^2$

Now, we can solve this equation for the time, $t_{1}$. First, multiply both sides by 2 and divide by $g$:

$t_{1}^2 = \frac{2 \cdot h}{g}$

Finally, take the square root of both sides to get the expression for $t_{1}$:

$t_{1} = \sqrt{\frac{2 \cdot h}{g}}$