Question

A ball is dropped into a well in whihthe water level is at the depth h below the top.If the speed if sound is v then calculate the time after which the splash is heared.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{for}\mathrm{the}\mathrm{ball}\mathrm{to}\mathrm{hit}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{water}\mathrm{be}{\mathrm{t}}_1\mathrm{seconds}. \\ \mathrm{h}=\frac{1}{2}{{\mathrm{gt}}_1}^2 \\ {\mathrm{t}}_1=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} \\ \mathrm{Let}\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{for}\mathrm{the}\mathrm{sound}\mathrm{to}\mathrm{be}\mathrm{heard}\mathrm{after}\mathrm{the}\mathrm{ball}\mathrm{hits}\mathrm{the}\mathrm{surface}\mathrm{be}{\mathrm{t}}_2 \\ {\mathrm{t}}_2=\frac{\mathrm{h}}{\mathrm{v}} \\ \mathrm{Total}\mathrm{time}\mathrm{taken}\mathrm{for}\mathrm{the}\mathrm{sound}\mathrm{to}\mathrm{be}\mathrm{heard}\mathrm{after}\mathrm{it}\mathrm{is}\mathrm{dropped}, \\ \mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2 \\ \mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}+\frac{\mathrm{h}}{\mathrm{v}}\mathrm{seconds} \end{gathered}$$