Inclined Terminal.

Let first define the Cartesian Axes as follow:

Also let the velocity of the particle be $u$ at any instant and makes an angle $\alpha$ with X-axis as shown.

Also let the velocity of particle be $u_{x}$ & $u_{y}$ along X-axis & Y-axis respectively at that instant.

Now draw the free body diagram of the particle. It will be like as below. (I have not mention the normal reaction on the particle. It will be perpendicular and outward to our defined axes).

Here f is the magnitude of frictional force acting on the particle i.e. $\mu mgcos\theta$ or $mgsin\theta$ . This means when the particle is traveling only down the inclined plane, its velocity will be constant. So at that moment $u_{y} = 0$

Now along X-axis we can write

$mgsin\theta - fcos\alpha = m\frac{du_{x}}{dt}$

And along Y-axis we can write

$fsin\alpha = -m\frac{du_{y}}{dt}$

Dividing both equations we get.

$\frac{mgsin\theta - fcos\alpha}{fsin\alpha} = -\frac{du_{x}}{du_{y}}$

$\implies$ $cosec\alpha - cot\alpha = -\frac{du_{x}}{du_{y}}$

$\implies$ $\frac{\sqrt{{u_{x}}^{2} + {u_{y}}^{2}} - u_{x}}{u_{y}} = -\frac{du_{x}}{du_{y}}$

$\implies$ $\sqrt{{u_{x}}^{2} + {u_{y}}^{2}}du_{y} = u_{x}du_{y} - u_{y}du_{x}$

The solution of this differential equation is $(u_{y})^{2} = c^2 - 2cu_{x}$ where $c$ is a constant.

When $u_{x} = 0$ then $u_{y} = v$ . This gives $c = v$ .

Now when $u_{y} = 0$ then $0 = v^2 - 2vu_{x}$

So $\boxed{u_{x} = \frac{v}{2} = 4.5 ms^{-1}}$