Time taken by the projectile to reach from $A$ to $B$ is $t$. Then, the distance $AB$ is equal to

The correct option is A $\frac{ut}{\sqrt{3}}$

Angle of projection with the horizontal $={60}^{\circ }$
Horizontal distance covered in time $t$
$=u\cos {60}^{\circ }\times t=AB\cos {30}^{\circ }$ (by geometry)
$\therefore AB=\frac{ut}{\sqrt{3}}$