Question

A swimmer sees an object that falls on the sea. He hears the sound of impact twice: once by the ear that is below the water surface and the other time by the ear that is in the air, 1 second later. At what distance from the observer did the impact took place??

Answer 1

Answer on Question #65211, Physics / Mechanics | Relativity

Question:

A swimmer sees an object that falls on the sea. He hears the sound of impact twice: once by the ear that is below the water surface and the other time by the ear that is in the air, 1 second later. At what distance from the observer did the impact took place?

Solution:

Let $L$ be the distance from the swimmer to the impact,

$v_{air}$ — speed of sound in the air,

$v_{water}$ — speed of sound in sea water,

$\Delta t$ — delay between the two events of detecting the sound.

We may write that $\Delta t = \frac{L}{v_{air}} - \frac{L}{v_{water}} = \frac{L(v_{water} - v_{air})}{v_{water} \cdot v_{air}}$, and then $L = \frac{v_{water} \cdot v_{air} \cdot \Delta t}{v_{water} - v_{air}}$.

$v_{air} = 331 \, \text{m/s}$

$v_{water} \cong 1510 \, \text{m/s}$

$\Delta t = 1 \, \text{s}$

$L = \frac{1510 \cdot 331 \cdot 1}{1510 - 331} \cong 424 \, \text{m}$

Answer:

424 m

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