The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is  $\sqrt{3}/2u$. The time of flight of the projectile is 

During the projectile motion, horizontal component of velocity does not change.
Given$u\cos \theta =\frac{\sqrt{3}}{2}u\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}$ that,  $v_x=\frac{\sqrt{3}}{2}u$
   $\theta ={30}^0$
 Time of flight.
$T=\frac{2u\sin \theta }{g}=\frac{2u\sin {30}^0}{g}$  =$\frac{2u}{g}$   x  $\frac{1}{2}$
 $\Rightarrow T=\frac{u}{g}$