Question

A stone is dropped in a well of depth$19.6m$. After how many seconds the sound of splash will be heard to the observer.velocity of sound in $\mathrm{air}=340{\mathrm{ms}}^{-1}$

Answer 1

Step1:- Given

Depth of well$=19.6\mathrm{m}$

Velocity of sound$=340\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step2:- Find the time taken by the stone to fall the distance$19.6\mathrm{m}$ is ${\mathrm{t}}_1$

By using newton's law of motion

$\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$

When,

$$\begin{gathered} \mathrm{u}=\mathrm{initial}\mathrm{velocity}=0 \\ \mathrm{g}=\mathrm{gravitational}\mathrm{acceleration}=9.8\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right. \\ \mathrm{h}=\mathrm{height} \\ \mathrm{t}=\mathrm{time} \end{gathered}$$

$$\begin{gathered} \therefore h=\frac{1}{2}g{t_1}^2 \\ {t}_1=\sqrt{\frac{2h}{g}} \\ =\sqrt{\frac{2\times 19.6}{9.8}} \\ {t}_2=2sec \end{gathered}$$

Step3:- Time taken by the sound of a splash to come up is ${\mathrm{t}}_2$

$$\begin{gathered} \mathrm{Time}=\frac{\mathrm{distance}}{\mathrm{velocity}} \\ {\mathrm{t}}_2=\frac{19.6}{340} \\ {\mathrm{t}}_2=0.05\sec \end{gathered}$$

Hence, the total time$={\mathrm{t}}_1+{\mathrm{t}}_2$

$$\begin{gathered} =2+0.05 \\ =2.05sec \end{gathered}$$

Hence the sound of splash will be heard after 2.05 seconds.