The sonar pulse takes t=4.2 s to reach the ocean bottom and then to travel back tothe ship. If we call d the depth of the ocean bottom, this means that the sonar pulse covered a distance equal to 2d in a time of 4.2 s.
The speed of the sonar pulse is the speed of sound in water at 25C, so looking at the table is v=1493 m/s. Using the relationship [tex]v= \frac{S}{t} [/tex] between space and time for a uniform motion, and keeping in mind that the total distance covered by the pulse is 2d, we can write:
[tex]v= \frac{2d}{t} [/tex]
which becomes
[tex]d= \frac{vt}{2}= \frac{(1493 m/s)(4.2 s)}{2}=3135 m [/tex]
so, the ocean bottom is 3135 m deep.
