$1.4.7^*.$ What is the duration of a plane's flight from Novosibirsk to Moscow and back in a straight line, if the wind blows at an angle $\alpha$ to the track at a speed $u$ during the entire flight? The speed of the aircraft relative to the air $v$, the length of the route $L$. In which wind direction is the maximum flight duration?
For the plane to fly on course, the following conditions must be met
$$u \sin \alpha = v \sin \beta$$
Where from
$$\cos \beta = \sqrt{1 - u^2 \sin ^2 \alpha / v^2}$$
And the total time there and back
$$ t_1 = \frac{L}{v \cos \beta + u \cos \alpha} $$
$$ t_2 = \frac{L}{v\cos \beta - u \cos \alpha} $$
We find the full time as
$$t=t_1+t_2$$
Substitute the value of $\cos \beta$:
$$ t=\frac{L}{\sqrt{v^2 - u^2 \sin ^2 \alpha } + u \cos \alpha} + \frac{L}{\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha} $$
$$ t=L\frac{\sqrt{v^2 - u^2 \sin ^2 \alpha }+\sqrt{v^2 - u^2 \sin ^2 \alpha} }{(\sqrt{v^2 - u^2 \sin ^2 \alpha} + u \cos \alpha)(\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha)} $$
$$ t=\frac{2L\sqrt{v^2 - u^2 \sin ^2 \alpha }}{(\sqrt{v^2 - u^2 \sin ^2 \alpha} + u \cos \alpha)(\sqrt{v^2 - u^2 \sin ^2 \alpha} - u \cos \alpha)} $$
Expressing the required time:
$$ \fbox{$t=\frac{2L\sqrt{v^2 - u^2 \sin ^2 \alpha }}{v^{2}-u^{2}}$} $$
$$t=\frac{2L\sqrt{v^{2}-u^{2}\operatorname{sin}^{2}\alpha}}{v^{2}-u^{2}}.\textrm{ Along the highway}.$$