Question:A submerged scuba diver hears the sound of a boat horn directly above her on the surface of the lake. At the same time, a friend on dry land 22.0 m from the boat also hears the horn (Fig. E16.7). The horn is 1.2 m above the surface of the water. What is the distance (labeled “?”) from the horn to the diver? Both air and water are at 20° C.

The given data can be listed below as,

• The distance between the friend and the boat is,\({d_1} = 22\;{\rm{m}}\).
• The distance between the horn and the surface is,\({d_2} = 1.2\;{\rm{m}}\).
• Both air and water are at 20° C.

The horn is the source here that generates the sound waves that reach the friend while it is traveling in the medium air, and it travels in another medium water and reaches the scuba diver, which is under the water as it is necessary for sound waves to have a medium they can travel in.

By using the given distances and the velocities, we can calculate the distance of the scuba diver from the surface.

The time taken by the sound wave to travel the distance from a submerged scuba diver to the horn and the time taken by the sound wave to travel the distance from the friend to the horn are the same.

The velocity of the sound wave at 20° C in the air is\({v_{air}} = 344\;{\rm{m/s}}\)and in water is\({v_{{\rm{water}}}} = 1482\;{\rm{m/s}}\).

We can calculate the time taken by sound waves in the air as,

\(t = \frac{{{d_1}}}{{{v_{air}}}}\) (1)

We can calculate the time taken by sound waves in the water as,

\(t = \frac{{{d_2}}}{{{v_{air}}}} + \frac{h}{{{v_{water}}}}\) (2)

Here\(h\)is the distance of the scuba diver from the surface.

Equating equations 1 and 2 as,

\(\begin{array}{c}\frac{{{d_2}}}{{{v_{air}}}} + \frac{h}{{{v_{water}}}} = \frac{{{d_1}}}{{{v_{air}}}}\\\frac{h}{{{v_{water}}}} = \frac{{{d_1}}}{{{v_{air}}}} - \frac{{{d_2}}}{{{v_{air}}}}\\h = {v_{water}}\left( {\frac{{{d_1}}}{{{v_{air}}}} - \frac{{{d_2}}}{{{v_{air}}}}} \right)\end{array}\)

Substitute the values in the above expression, and we get,

\(\begin{array}{c}h = 1482\;{\rm{m/s}}\left( {\frac{{22\;{\rm{m}}}}{{344\;{\rm{m/s}}}} - \frac{{1.2\;{\rm{m}}}}{{344\;{\rm{m/s}}}}} \right)\\ = 4.308\left( {22\;{\rm{m}} - 1.2\;{\rm{m}}} \right)\\ = 4.308\left( {20.8\;{\rm{m}}} \right)\\ = {\rm{89}}{\rm{.60}}\;{\rm{m}}\end{array}\)

Now the distance from horn to scuba diver is the sum of the distance oh horn from the surface and the distance of diver from the surafce; we can calculate it as,

\(D = h + {d_2}\)

Substitute the values in the above expression, and we get,

\(\begin{array}{c}D = {\rm{89}}{\rm{.60}}\;{\rm{m}} + {\rm{1}}{\rm{.2}}\;{\rm{m}}\\ = 90.8\;{\rm{m}}\end{array}\)

Thus, the distance (labeled “?”) from the horn to the diver is \(90.8\;{\rm{m}}\) .