Question Number 54584 by peter frank last updated on 07/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19
$${trying} \\ $$$${t}_{{echo}\:{from}\:{bottom}} =\frac{\mathrm{2}\left(\mathrm{200}\right)}{{v}_{{sound}\:{inair}} }+\frac{\mathrm{2}{h}}{{v}_{{sound}\:{in}\:{water}} } \\ $$$${t}_{{echo}\:{from}\:{surface}} =\frac{\mathrm{2}\left(\mathrm{200}\right)}{{v}_{{sound}\:{in}\:{air}} } \\ $$$${t}_{{b}} −{t}_{{s}} =\mathrm{0}.\mathrm{24}=\frac{\mathrm{2}{h}}{{v}_{{sound}\:{in}\:{water}} } \\ $$$${h}=\frac{\mathrm{0}.\mathrm{24}×{v}_{{sound}\:{in}\:{water}} }{\mathrm{2}} \\ $$
Commented by peter frank last updated on 07/Feb/19
$${thank}\:{you} \\ $$
Answered by mr W last updated on 07/Feb/19
$${t}_{\mathrm{1}} =\mathrm{2}{h}/{v}_{{air}} \\ $$$${t}_{\mathrm{2}} =\mathrm{2}{h}/{v}_{{air}} +\mathrm{2}{d}/{v}_{{water}} \\ $$$$\Delta{t}={t}_{\mathrm{2}} −{t}_{\mathrm{1}} =\mathrm{2}{d}/{v}_{{water}} \\ $$$$\Rightarrow{d}=\frac{\Delta{t}\:{v}_{{water}} }{\mathrm{2}}=\frac{\mathrm{0}.\mathrm{24}×\mathrm{1481}}{\mathrm{2}}=\mathrm{178}{m} \\ $$
Commented by mr W last updated on 07/Feb/19
