To solve this problem it is necessary to resort to concepts related to the kinematic equations of movement description which define speed as
[tex]v = \frac{x}{t}[/tex]
Where,
x = Distance
t=Time
Re-arrange to find the time we have,
[tex]t = \frac{x}{v}[/tex]
At 20°C the speed of sound in air is 343.2m/s and the speed of sound in water is 1484.3m/s
The time in which the sound lasts in traveling in the air would be given by the shipping and return (even when it touches the water) which is 28m (14 way and 14 back), in other words
[tex]t_a = \frac{28}{343.2} = 0.08158s[/tex]
In the case of the lake, the same logic is followed, only that there is no net distance here, therefore
[tex]t_w = \frac{2d}{1484.3}[/tex]
The total time would be given by
[tex]t_T = 0.08158 + \frac{2d}{1484.3}[/tex]
The total time is given in the statement so we clear to find d
[tex]0.108-0.08158= \frac{2d}{1484.3}[/tex]
[tex]d = \frac{1484.3(0.108-0.08158)}{2}[/tex]
[tex]d = 18.86m[/tex]
Therefore the lake has a depth of 18.86 m
