13.128. Above point $A$, the helicopter was at 8:30 AM. After flying a straight distance of $s$ km, the helicopter arrived at point $B$. Hovering over point $B$ for 5 minutes, the helicopter then returned along the same route. It returned to point $A$ at 10:35 AM. From $A$ to $B$, it flew with the wind, and on the return trip, it flew against the wind. The wind speed was constant throughout. Find the wind speed, given that the helicopter's own speed was also constant and equal to $\mathrm{v}$ km/h in calm conditions:

Under what relationship between the given quantities does the problem have a solution
Solution.

Let $x$ hours be the time taken to travel from $A$ to $B$, and $y$ hours from $B$ to $A$. According to the condition, $x+y=2$. The speed of the helicopter from $A$ to $B$ was $v+v_{\mathrm{B}}=\frac{s}{x}$, and on the return trip - $v-v_{\mathrm{B}}=\frac{s}{y}$. Solving the system $\left\{\begin{array}{l}x+y=2, \\ v+v_{\mathrm{B}}=\frac{s}{x}, \text { we find } \\ v-v_{\mathrm{B}}=\frac{s}{y},\end{array}\right.$ $v_{\mathrm{B}}=\sqrt{v(v-s)}$ km/h for $v>s$.

Answer: $\sqrt{v(v-s)}$ km/h for $v>s$.