Question: A stone is dropped into a well. The splash is heard 3.00s later. What is the depth of the well?

The total time is t = 3.00s

First, find the equation for thetotal timebetween the dropping of the stone to hear the splash. Then, reduce this equation to the quadratic equation of. Finally, using thequadratic solution forand substituting the given values, find the depthof the well.

The total time interval between the dropping of stone to hearing the splash is,

.

Let, t_f be the time for the stone to fall into the water and t_s be the time for the sound of the splash to travel from the water to the top of the well. Thus, the total time elapsed from the dropping of the stone to hearing the splash is,

$t=t_f+t_s$

As, the initial velocity of the stone is zero, Ifis the depth of the well, then the kinematics of free fall gives,

$d=\frac{1}{2}g{t_f}^2$

Therefore, the time t_f is given by,

$t_f=\sqrt{\frac{2d}{g}}$

Let the speed of sound in air be v_s, so,

$d=v_s{t}_s$

Thus, the total time is given by,

$t=\sqrt{\frac{2d}{g}}+\frac{d}{v_s}$

Rewrite it as,

$\sqrt{\frac{2d}{g}}=t-\frac{d}{v_s}$

And squaring both sides to obtain,

$\frac{2d}{g}=t^2-2\left(\frac{t}{v_s}\right)d+\left(1+{v_s}^2\right)d^2$

Now, multiplying by$g{t_s}^2$,

$g{t_s}^2\times \frac{2d}{g}=g{t_s}^2{t}^2-2\left(\frac{t}{v_s}\right)g{t_s}^{2}d+\left(1+{v_s}^2\right)g{t_s}^2{d}^2$

Rearrange to get,

$g{d}^2-2v_s\left(gt+v_s\right)d+g{v_s}^2{t}^2=0$

This is a quadratic equation for d.

Therefore, the solution is,

$d=\frac{2v_s\left(gt+v_s\right)\pm \sqrt{4{v_s}^2{\left(gt+v_s\right)}^2-4g^2{v_s}^2{t}^2}}{2g}$

As distance cannot be negative, the only solution of the quadratic equation is-

$d=\frac{2v_s\left(gt+v_s\right)-\sqrt{4{v_s}^2{\left(gt+v_s\right)}^2-4g^2{v_s}^2{t}^2}}{2g}$

By substituting ,$g={\text{9.8m/s}}^{\text{2}}\text{,}{v}_s=\text{343m/sand}t=3.\text{00s}$

$$\begin{gathered} d=\left(\frac{2\times \left(343\text{m/s}\right)\left(9.8{\text{m/s}}^{\text{2}}\times \left(3.00\text{s}\right)+\left(343\text{m/s}\right)\right) \\ -4\times {\left(343\text{m/s}\right)}^2{\left(9.8{\text{m/s}}^{\text{2}}\times \left(3.00\text{s}\right)+\left(343\text{m/s}\right)\right)}^2 \\ -4\times {\left(9.8{\text{m/s}}^{\text{2}}\right)}^2\times {\left(343\text{m/s}\right)}^2\times {\left(3.00\text{s}\right)}^2 \\ }{2\times 9.8{\text{m/s}}^{\text{2}}}\right) \end{gathered}$$

d = 40.7

Hence,the depth of the well is 40.7.