Hunting Red October with echolocation

The secondary signal is the echo reflected from the water surface. Therefore, the signals corresponds to different path length $s$ and $s' = \sqrt{s^2 + 4 d^2}$ and arrive at different times $t$ and $t'$ at the hydrophone with $\begin{aligned} & & t &= \frac{s}{c} + t_0 \\ & & t' &= \frac{\sqrt{s^2 + 4 d^2}}{c} + t_0 \\ \Rightarrow & & c(t' - t) &= \sqrt{s^2 + 4 d^2} - s \\ \Rightarrow & & (c(t'- t) + s)^2 &= s^2 + 4 d^2 \\ \Rightarrow & & c^2(t'- t)^2 + 2 s c(t'- t) &= 4 d^2 \\ \Rightarrow & & s &= \frac{4 d^2 - \delta^2}{2 \delta} \end{aligned}$ with the path difference $\delta = c(t' - t)$ . Therefore, we can estimate the distances $\begin{aligned} s_{A,1} &\approx 840.8\,\text{m}\\ s_{B,1} &\approx 850.1\,\text{m}\\ s_{A,2} &\approx 689.0\,\text{m}\\ s_{B,2} &\approx 646.8\,\text{m} \end{aligned}$ between the hydrophones $A$ and $B$ and the sound source. With the help of triangulation we can estimate the ( $x,y$ ) coordinates of the enemy submarine: $\begin{aligned} & & s_A^2 &= x^2 + y^2 \\ & & s_B^2 & = (x - l)^2 + y^2 \\ \Rightarrow & & s_A^2 - s_B^2 &= 2 x l - l^2 \\ \Rightarrow & & x &= \frac{ s_A^2 - s_B^2 + l^2}{2l} \\ \Rightarrow & & y &= \sqrt{s_A^2 - x^2} \end{aligned}$ Numerical evalution results the coordinates $\begin{aligned} (x_1,y_1) &\approx (4, 841) \,\text{m} \\ (x_2,y_2) &\approx (282, 629) \,\text{m} \end{aligned}$ With a time difference of $\Delta t \approx 50 \,\text{s}$ between both events we can estimate a velocity $v = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} }{\Delta t} \approx 6.99 \,\frac{\text{m}}{\text{s}}$