**Problem 1005**

A stone is dropped down a well and 5 sec later, the sounds of the splash is heard. If the velocity of sound is 1120 ft/sec (341.376 m/s), what is the depth of the well?

**Solution**

_{stone}(free-falling body):

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

$t = \sqrt{\dfrac{2h}{g}}$

For t_{sound} (uniform motion):

$h = vt$

$t = \dfrac{h}{v}$

Thus,

$\sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} = 5$

English System

$\sqrt{\dfrac{2h}{32.2}} = 5 - \dfrac{h}{1120}$

$\dfrac{2h}{32.2} = \left( 5 - \dfrac{h}{1120} \right)^2$

$\dfrac{10h}{161} = 25 - \dfrac{h}{112} + \dfrac{h^2}{1\,254\,400}$

$\dfrac{1}{1\,254\,400}h^2 - \dfrac{183}{2576}h + 25 = 0$

$h = 88\,759.73 \, \text{ and } \, 353.31$

For h = 88 759.73 ft

$t = \sqrt{\dfrac{2(88\,759.73)}{32.2}}$

$t = 74.2 ~ \text{sec} ~ \gt 5 ~ \text{sec}$ (*not okay!*)

For h = 353.31 ft

$t = \sqrt{\dfrac{2(353.31)}{32.2}}$

$t = 4.68 ~ \text{sec} ~ \lt 5 ~ \text{sec}$ (*okay!*)

Thus, h = 353.31 ft → *answer*

SI Units

$\sqrt{\dfrac{2h}{9.81}} = 5 - \dfrac{h}{341.376}$

$\sqrt{\dfrac{200h}{981}} = 5 - \dfrac{125h}{42\,672}$

$\dfrac{200h}{981} = \left( 5 - \dfrac{125h}{42\,672} \right)^2$

$\dfrac{200h}{981} = 25 - \dfrac{625h}{21\,336} + \dfrac{15\,625h^2}{1\,820\,899\,584}$

$\dfrac{15\,625}{1\,820\,899\,584}h^2 - \dfrac{1\,626\,775}{6\,976\,872}h + 25 = 0$

$h = 27\,065.05 \, \text{ and } \, 107.64$

For h = 27 065.05 m

$t = \sqrt{\dfrac{2(27\,065.05)}{9.81}}$

$t = 74.2 ~ \text{sec} ~ \gt 5 ~ \text{sec}$ (*not okay!*)

For h = 107.64 m

$t = \sqrt{\dfrac{2(107.64)}{9.81}}$

$t = 4.68 ~ \text{sec} \lt 5 ~ \text{sec}$ (*okay!*)

Thus, h = 107.64 m → *answer*