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

The correct option is A $\frac{ut}{\sqrt{3}}$
Let the projection of AB on the horizontal axis be $d$.
Then,
$d=u\cos {60}^{\circ }t$
$AB\cos {30}^{\circ }=u\cos {60}^{\circ }t$
$\cos {30}^{\circ }=\frac{u\cos {60}^{\circ }t}{AB}$
$AB=u\times \frac{1}{2}\times \frac{2}{\sqrt{3}}\times t$
$AB=\frac{ut}{\sqrt{3}}$