## College Physics (4th Edition)

The depth of the ocean at that location is $~5,419~m$
We can use $v = 1533~m/s$ as the speed of sound in seawater at a temperature of $25^{\circ}C$. We can find the total distance the sound travels in the water: $d = v~t = (1533~m/s)(7.07~s) = 10,838~m$ Since the sound reached the bottom of the ocean and returned to the ship, the depth of the water is $\frac{10,838~m}{2} = 5,419~m$