1. Resolve the initial velocity

Let the projectile be fired with speed $u$ at an angle $\theta$ to the horizontal.

So, $u_x = u\cos\theta, \qquad u_y = u\sin\theta$

2. Use the condition at the highest point

At the highest point, the vertical component of velocity becomes zero, so the speed there is only the horizontal component: $v_{\text{top}} = u\cos\theta$

Given: $u\cos\theta = \frac{\sqrt{3}}{2}u$

Cancelling $u$: $\cos\theta = \frac{\sqrt{3}}{2}$

Thus, $\theta = 30^\circ$

Hence, $\sin\theta = \frac{1}{2}$

3. Formula for time of flight

For a projectile fired from ground and landing back on ground, $T = \frac{2u\sin\theta}{g}$

Substitute $\sin\theta = \frac{1}{2}$: $T = \frac{2u\left(\frac{1}{2}\right)}{g} = \frac{u}{g}$

4. Match with the options

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

So the correct option is A.