# A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00s later. Find the depth of the well if the air temperature is 10.0°c ?

Aug 26, 2015

$674$ meters

#### Explanation:

Although there will be some variation depending upon air humidity (for example) the speed of sound relative to temperature (measured in degrees Celsius) can be determined by the empirical formula:
color(white)("XXX")s=331" meters"/"sec."+0.6 "meters"/"sec."*c
where $c$ is the temperature in degrees Celsius.

For a speed of $s$ and time of $t$,
the distance $d$ is
color(white)(*XXX")d = sxxt

Therefore we have
color(white)("XXX")D =331+0.6(10) "meters"/"second" xx 2 "seconds"

color(white)("XXXX")= 674 "meters"

Aug 27, 2015

$19.8 \text{m}$

#### Explanation:

For the 1st part the stone is falling under gravity. Let ${t}_{1}$ be the time from release to splashdown.

$d$ =depth

So $d = \frac{1}{2} \text{g} {t}_{1}^{2}$ $\text{ }$$\textcolor{red}{\left(1\right)}$

After splashdown the sound travels back up. Using $330 \text{m/s}$ for the speed of sound this gives:

$d = 330 \times {t}_{2}$$\text{ }$ $\textcolor{red}{\left(2\right)}$

We also know that:

${t}_{1} + {t}_{2} = 2 \text{s}$ $\text{ } \textcolor{red}{\left(3\right)}$

Combining $\textcolor{red}{\left(1\right)} \text{and} \textcolor{red}{\left(2\right)}$ we get:

$330 {t}_{2} = \frac{1}{2} g . {t}_{1}^{2}$

From $\textcolor{red}{\left(3\right)} \Rightarrow$

${t}_{2} = \left(2 - {t}_{1}\right)$

So:

$330 \left(2 - {t}_{1}\right) = \frac{1}{2.} \text{g} {t}_{1}^{2}$

$660 - 330 {t}_{1} = \frac{1}{2.} \text{g} {t}_{1}^{2}$

$\frac{1}{2} \text{g} {t}_{1}^{2} + 330 {t}_{1} - 660 = 0$

Using the quadratic formula to solve for ${t}_{1}$:

${t}_{1} = \frac{- 330 \pm \sqrt{{\left(330\right)}^{2} - 4 \times \frac{9.8}{2} \times \left(- 660\right)}}{9.8}$

Ignoring the -ve root this gives:

${t}_{1} = 1.94 \text{s}$

So ${t}_{2} = 2 - 1.94 = 0.06 \text{s}$

So $d = 330 \times {t}_{2} = 330 \times 0.06 = 19.8 \text{m}$