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

For a projectile motion, we know 

At highest point, vertical component of velocity becomes zero.

Thus, there exists only horizontal component of velocity, ${\mathrm{u}}_{\mathrm{y}}=\mathrm{ucos\theta }$

Given

$\mathrm{ucos\theta }=\frac{\sqrt{3}}{2}\mathrm{u}$

$\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{6}$

We know that in a projectile motion, Time of flight is given by 

$\mathrm{T} =\frac{\mathrm{usin\theta }}{\mathrm{g}}$

$\therefore \mathrm{T}=\frac{\mathrm{u}}{\mathrm{g}}$