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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) =((2(200))/v_(sound inair) )+((2h)/v_(sound in water) )  t_(echo from surface) =((2(200))/v_(sound in air) )  t_b −t_s =0.24=((2h)/v_(sound in water) )  h=((0.24×v_(sound in water) )/2)

$${trying} \\ $$$${t}_{{echo}\:{from}\:{bottom}} =\frac{\mathrm{2}\left(\mathrm{200}\right)}{{v}_{{sound}\:{inair}} }+\frac{\mathrm{2}{h}}{{v}_{{sound}\:{in}\:{water}} } \\ $$$${t}_{{echo}\:{from}\:{surface}} =\frac{\mathrm{2}\left(\mathrm{200}\right)}{{v}_{{sound}\:{in}\:{air}} } \\ $$$${t}_{{b}} −{t}_{{s}} =\mathrm{0}.\mathrm{24}=\frac{\mathrm{2}{h}}{{v}_{{sound}\:{in}\:{water}} } \\ $$$${h}=\frac{\mathrm{0}.\mathrm{24}×{v}_{{sound}\:{in}\:{water}} }{\mathrm{2}} \\ $$

Commented by peter frank last updated on 07/Feb/19

thank you

$${thank}\:{you} \\ $$

Answered by mr W last updated on 07/Feb/19

t_1 =2h/v_(air)   t_2 =2h/v_(air) +2d/v_(water)   Δt=t_2 −t_1 =2d/v_(water)   ⇒d=((Δt v_(water) )/2)=((0.24×1481)/2)=178m

$${t}_{\mathrm{1}} =\mathrm{2}{h}/{v}_{{air}} \\ $$$${t}_{\mathrm{2}} =\mathrm{2}{h}/{v}_{{air}} +\mathrm{2}{d}/{v}_{{water}} \\ $$$$\Delta{t}={t}_{\mathrm{2}} −{t}_{\mathrm{1}} =\mathrm{2}{d}/{v}_{{water}} \\ $$$$\Rightarrow{d}=\frac{\Delta{t}\:{v}_{{water}} }{\mathrm{2}}=\frac{\mathrm{0}.\mathrm{24}×\mathrm{1481}}{\mathrm{2}}=\mathrm{178}{m} \\ $$

Commented by mr W last updated on 07/Feb/19

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