## Statement

Determine the depth of a well in which a stone is dropped, and where you can hear the impact on the water after 1.5 s, considering that the velocity of sound is 340 m/s.

## Solution

**Data**

t = 1.5 s

v_{sound} = 340 m/s

g = 9.8 m/s2

H = ?

Resolution

Let's consider the exercise in two parts:

- Part 1. The stone descends to impact the water at the bottom of the well
- Part 2. The sound moves up from the water to the opening of the well

If we call the time the stone takes to get to the water t_{1} and t_{2} the time the sound takes to get back to the opening of the well, we get that:

$$t={t}_{1}+{t}_{2}$$

We will study each part separately:

Part 1

Applying the equation of position of the fall free motion and knowing that at t_{1}, the position of the stone is y = 0 m:

$$y=H-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}0=H-0.5\xb79.8\xb7{{t}_{1}}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)H=4.9\xb7{{t}_{1}}^{2}}$$

Part 2

Sound rises at constant speed, and we assume that it does it in a straight line. Therefore:

$$x=v\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}H={v}_{sonido}\xb7{t}_{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)H=340\xb7{t}_{2}}$$

Since H is the same in both equations and we know that t=t_{1}+t_{2}:

$$\left.\begin{array}{r}4.9\xb7{{t}_{1}}^{2}=340\xb7{t}_{2}\\ 1.5={t}_{1}+{t}_{2}\end{array}\right\}\Rightarrow \phantom{\rule{0ex}{0ex}}\left.\begin{array}{r}4.9\xb7{{t}_{1}}^{2}-340\xb7{t}_{2}=0\\ {t}_{2}=1.5-{t}_{1}\end{array}\right\}\Rightarrow \phantom{\rule{0ex}{0ex}}4.9\xb7{{t}_{1}}^{2}-340\xb7\left(1.5\xb7{t}_{1}\right)=0\Rightarrow \phantom{\rule{0ex}{0ex}}4.9\xb7{{t}_{1}}^{2}-340\xb7{t}_{1}-510=0\Rightarrow \phantom{\rule{0ex}{0ex}}{t}_{1}=\frac{-340\pm \sqrt{{340}^{2}-1\xb74.9\xb7(-510)}}{2\xb74.9}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){t}_{1}=1.47s}\phantom{\rule{0ex}{0ex}}$$

Substituting t_{1} in the equation of part 1:

$$H=4.9\xb7{\left(1.47\right)}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)H=10.59m}$$