A submarine on the ocean surface gets a sonar echo indicating an underwater object.

Question

The echo comes back at an angle of 20 degrees above the horizontal and the echo took 2.32 s to get back to the submarine. What is the object's depth?

Answer 1

First you need to know the average speed of sound for the water column (Speed of Sound varies in every part of the ocean). Once you have determined the speed of sound do this.

(Delta t x Speed of Sound) / 2 = Straight distance to the object

sin of 20 x straight line distance = depth

i,e assuming the average SOS is 1500m/s

(1500m/s x 2.32) / 2 = 1740m to the object

sin 20 (.342) x 1740m = 595.08m

Answer 2

The distance to the target is the speed of sound times half the echo time (since it had to get there and back).

distance = ct/2

Finding the depth is just simple trig.  Draw out the triangles and see that:

depth = distance * sin (angle)

= sin(angle) c t / 2

They give you the time and the angle.  You'll need to look up the speed of sound in water.  Plugnchug.