Answer:
Approximately [tex]2.9 \times 10^2 \; \rm m[/tex].
Explanation:
In that [tex]0.39\; \rm s[/tex], this sound wave travelled from the surface of the lake to the bottom, got reflected, and travelled back from the bottom to the surface. The sound wave travelled from the surface to the bottom (without bouncing back) in only [tex]1/2[/tex] that much time. In other words, it took only [tex]\displaystyle ((1/2) \times 0.39)\; \rm s[/tex] for the sound wave to travel from the surface to the bottom of the lake.
The speed [tex]v[/tex] of sound in cold water ([tex]20\; \rm ^\circ C[/tex], [tex]1\; \rm atm[/tex]) is approximately [tex]1.482\times 10^{3} \; \rm m \cdot s^{-1}[/tex].
In [tex]t = ((1/2) \times 0.39) \; \rm s[/tex], that sound wave would have travelled a distance of:
[tex]\begin{aligned}s &= v \cdot t \\ &= 1.482 \times 10^3 \; \rm m \cdot s^{-1} \times \left(\frac{1}{2} \times 0.39 \; \rm s\right) \\ &\approx 2.9 \times 10^2\; \rm m \end{aligned}[/tex].
Therefore, the depth of the lake is approximately [tex]2.9 \times 10^2 \; \rm m[/tex].
