Answer to Question #189711 in Mechanics | Relativity for Fhulu

Question

You drop a stone into a deep well and hear it hit the bottom 3.2 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound to reach you. Calculate how deep the well is. Sound travels about 343 m/s in air.

Answer 1

There are two parts in the question :-

$1.$ The stone dropped with zero initial velocity and it moves under gravity with constant acceleration$= g= 10 \frac{m}{sec^2}$

$2.$ The sound produced by stone travels back to the top surface.

Given Information :

Speed of Sound$=$ $343 \frac{m}{sec}$ $=V$

Time after dropping the stone when sound heard $=3.2 sec$

(it is the time taken by stone to reach bottom of the well then sound is produced and time taken by sound to move from bottom of the well to top )

Need to calculate = depth of well ?

Lets assume depth of well is common for both parts $=H$

so total time $T= t_{1} + t_{2}$ $.......(1)$

$t_{1} =$ time taken by stone to fall to bottom of well $= \sqrt\frac{2H}{g}$ $........(2)$

$t_{2} =$ time taken by sound to reach you $= \frac{H}{V}$ because speed of sound is constant $.........(3)$

Put the values of $t_{1}$ and $t_{2}$ from equation $(2)$ and $(3)$

$3.2 sec =$ $= \sqrt\frac{2H}{g}$ $+$ $\frac{H}{V}$

$\implies$ $3.2 - \frac{H}{V} = \sqrt{}{}\frac{2H}{g}$

Now after squaring both side ,

$(3.2-\frac{H}{V})^2$ $= \frac{2H}{g}$

$\implies 10.24+\frac{H^2}{V^2}$ $-6.4\frac{H}{V}$ $= \frac{2H}{g}$

$\implies \frac{H^2}{V^2} - \frac{6.4H}{V} - \frac{2H}{g}+10.24 = 0$

$\implies$ $H^2 - 21334.6H+1204725.76=0$

this can be solved by using quadratic formula

$\implies H = \frac{21334.6-\sqrt{21334.6^2-4 \times 1204725.76}}{2}$ (ignoring +ve root)

$\implies H =$ $\frac{21334.6-21221.3}{2}$ $=\frac{113.3}{2} = 56.65 m$

Answer : Well is 56.65m deep.