A man wants to reach point B on the opposite bank of a river flowing at a speed $u$ as shown. What minimum velocity relative to water should the man have so that he can reach directly to point B?

The correct option is A $\frac{u}{\sqrt{2}}$, in the upstream at an angle ${45}^{\circ }$ with the vertical
Let $v$ be the velocity of man w.r.t flow in a direction at an angle $\theta$ with the direction perpendicular to flow.
For the man to go along AB,
Along y-axis, $v\sin (\theta +{45}^{\circ })=u\sin {45}^{\circ }$
for him to go along AB, then
$v=\frac{u\sin {45}^{\circ }}{\sin (\theta +{45}^{\circ })}$
For $v$ to be minimum, $\sin (\theta +{45}^{\circ })$ should be maximum
$\Rightarrow \theta +{45}^{\circ }={90}^{\circ }$
$\Rightarrow \theta ={45}^{\circ }$
$\Rightarrow V_{min}=\frac{u}{\sqrt{2}}$, in the upstream at an angle ${45}^{\circ }$ to the vertical.