1. For motion up the incline

The body is projected upward on a rough incline of angle $\theta = 30^\circ$.

While moving upward, both gravity component and friction act down the plane.

So retardation during ascent is $a_1 = g\sin\theta + \mu g\cos\theta$

Since $\sin 30^\circ = \frac12$ and $\cos 30^\circ = \frac{\sqrt3}{2}$, $a_1 = g\left(\frac12 + \mu\frac{\sqrt3}{2}\right)$

If initial speed is $u$, then time of ascent is $t_1 = \frac{u}{a_1} = \frac{u}{g\left(\frac12 + \mu\frac{\sqrt3}{2}\right)}$

2. Distance travelled during ascent

At the highest point, final velocity becomes zero. Using $0 = u^2 - 2a_1 s$ we get $s = \frac{u^2}{2a_1}$

3. For motion down the incline

While sliding downward, gravity component acts down the plane and friction acts up the plane.

So acceleration during descent is $a_2 = g\sin\theta - \mu g\cos\theta$

Thus, $a_2 = g\left(\frac12 - \mu\frac{\sqrt3}{2}\right)$

The body starts from rest at the top and travels distance $s$ downward, so $s = \frac12 a_2 t_2^2$

Hence, $t_2 = \sqrt{\frac{2s}{a_2}}$

Substitute $s = \frac{u^2}{2a_1}$: $t_2 = \sqrt{\frac{2\left(\frac{u^2}{2a_1}\right)}{a_2}} = \frac{u}{\sqrt{a_1a_2}}$

4. Use the given condition

Given time of ascent is half of time of descent: $t_1 = \frac12 t_2$

Substitute expressions: $\frac{u}{a_1} = \frac12\cdot \frac{u}{\sqrt{a_1a_2}}$

Cancel $u$: $\frac{1}{a_1} = \frac{1}{2\sqrt{a_1a_2}}$

Cross-multiplying, $2\sqrt{a_1a_2} = a_1$

Squaring both sides: $4a_1a_2 = a_1^2$

Since $a_1 \neq 0$, $a_1 = 4a_2$

5. Substitute $a_1$ and $a_2$

$g\left(\frac12 + \mu\frac{\sqrt3}{2}\right) = 4g\left(\frac12 - \mu\frac{\sqrt3}{2}\right)$

Cancel $g$ and multiply by $2$: $1 + \mu\sqrt3 = 4(1 - \mu\sqrt3)$

$1 + \mu\sqrt3 = 4 - 4\mu\sqrt3$

$5\mu\sqrt3 = 3$

$\mu = \frac{3}{5\sqrt3} = \frac{\sqrt3}{5}$

6. Compare with the given form

Given $\mu = \frac{\sqrt{x}}{5}$

So, $\frac{\sqrt{x}}{5} = \frac{\sqrt3}{5}$

Therefore, $\sqrt{x} = \sqrt3$ $x = 3$

7. Final answer

$\boxed{3}$