Source: High School Physics Olympiad
Problem Sets:
Problem
In still air an airplane flies with uniform speed $v$ relative to the ground, and the farthest distance it can fly out (counting the trip out and back) is $R$. The airplane now flies in a wind of speed $u$ blowing toward the direction $\alpha$ degrees east of north, while the airplane's actual track over the ground is $\beta$ degrees east of north (out and back along the same line).
The total flight time available is $T = \dfrac{2R}{v}$.
Resolve the wind along and across the track. The angle between wind and track is $\alpha - \beta$, giving components $u_\parallel = u\cos(\alpha - \beta)$ and $u_\perp = u\sin(\alpha - \beta)$.
To stay on the track, the airspeed (magnitude $v$) must cancel $u_\perp$, leaving along-track airspeed $\sqrt{v^2 - u_\perp^2}$. The ground speeds out and back are:
$$v_1 = \sqrt{v^2 - u_\perp^2} + u_\parallel, \qquad v_2 = \sqrt{v^2 - u_\perp^2} - u_\parallel$$If $R'$ is the farthest distance, $\dfrac{R'}{v_1} + \dfrac{R'}{v_2} = T = \dfrac{2R}{v}$, so:
$$R' = \frac{2R}{v}\cdot\frac{v_1 v_2}{v_1 + v_2} = \frac{2R}{v}\cdot\frac{v^2 - u_\perp^2 - u_\parallel^2}{2\sqrt{v^2 - u_\perp^2}} = \frac{R(v^2 - u^2)}{v\sqrt{v^2 - u^2\sin^2(\alpha - \beta)}}$$