A projectile is thrown with a speed $u$ at an angle $\theta$ with the vertical. Its average velocity between the instants, it crosses half the maximum height is :

The correct option is B $u\sin \theta$, horizontal and in the plane of projection
Let ${\overrightarrow{V}}_1$ be velocity of particle at $A$
and ${\overrightarrow{V}}_2$ be velocity of particle at $B$

Now, let speed of projectile at half the max height be $V^{\prime },$
Horizontal component of velocity at each instant is $V_{1x}=u\cos \theta$ ($\because \theta$ is measured from vertical )
Here,
${\overrightarrow{V}}_1=V_{1x}\hat{i}+V_{iy}\hat{j}$ and ${\overrightarrow{V}}_2=V_{1x}\hat{i}-V_{iy}\hat{j}$
($\because$ both $A$ & $B$ are at same level)

Average velocity between $A$ & $B=\frac{{\overrightarrow{V}}_1+{\overrightarrow{V}}_2}{2}$
($\because$ acceleration is constant $=g$)
$\therefore \frac{{\overrightarrow{V}}_1+{\overrightarrow{V}}_2}{2}=V_{1x}\hat{i}=u\sin \theta \hat{i}$