A body of mass $$m$$ is launched up on a rough inclined plane making an angle of 30$$^\circ$$ with the horizontal. The coefficient of friction between the body and plane is $$\frac{\sqrt{x}}{5}$$ if the time of ascent is half of the time of descent. The value of $$x$$ is ___.

Let the coefficient of friction be $$\mu = \frac{\sqrt{x}}{5}$$ and the incline angle be $$\theta = 30^\circ$$. During ascent, both the component of gravity along the plane and friction act downward along the slope, so the deceleration is $$a_{\text{up}} = g(\sin\theta + \mu\cos\theta)$$. During descent, gravity acts down the slope while friction acts upward, giving acceleration $$a_{\text{down}} = g(\sin\theta - \mu\cos\theta)$$.

If the body is launched with initial speed $$u$$, it decelerates uniformly to rest, so the time of ascent is $$t_a = \frac{u}{a_{\text{up}}}$$. The distance traveled up the slope is $$s = \frac{u^2}{2a_{\text{up}}}$$. Starting from rest, the body descends this same distance $$s$$ with acceleration $$a_{\text{down}}$$, so $$s = \frac{1}{2}a_{\text{down}}t_d^2$$, giving $$t_d = \sqrt{\frac{2s}{a_{\text{down}}}} = \frac{u}{\sqrt{a_{\text{up}} \cdot a_{\text{down}}}}$$.

We are given $$t_a = \frac{1}{2}t_d$$:

$$\frac{u}{a_{\text{up}}} = \frac{1}{2} \cdot \frac{u}{\sqrt{a_{\text{up}} \cdot a_{\text{down}}}}$$

$$2\sqrt{a_{\text{up}} \cdot a_{\text{down}}} = a_{\text{up}}$$

Squaring both sides: $$4 a_{\text{down}} = a_{\text{up}}$$

Substituting with $$\sin 30^\circ = \frac{1}{2}$$ and $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$:

$$4\left(\frac{1}{2} - \mu\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \mu\frac{\sqrt{3}}{2}$$

$$2 - 2\sqrt{3}\mu = \frac{1}{2} + \frac{\sqrt{3}}{2}\mu$$

$$\frac{3}{2} = 2\sqrt{3}\mu + \frac{\sqrt{3}}{2}\mu = \frac{5\sqrt{3}}{2}\mu$$

$$\mu = \frac{3}{2} \cdot \frac{2}{5\sqrt{3}} = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$$

Since $$\mu = \frac{\sqrt{x}}{5}$$, we get $$\sqrt{x} = \sqrt{3}$$, so $$x = 3$$.