Terminal velocity of an object on an inclined plane?

Let x be the sideways direction, and let y be the down-hill direction. ( $V_x$ and $V_y$ are the components of velocity in those directions.)

The frictional force will be $F_f=mg\mu \cos\theta=mg\sin\theta$ ... Since friction points in the direction opposite to the motion, we have $\frac{F_{f.x}}{F_f}=\frac{V_x}{\sqrt{V_x^2+V_y^2}}$ and $\frac{F_{f.y}}{F_f}=\frac{V_y}{\sqrt{V_x^2+V_y^2}}$

Therefore the equations which describe the motion are:

$\frac{dV_x}{dt}=-g\sin\theta\frac{V_x}{\sqrt{V_x^2+V_y^2}}$

$\frac{dV_y}{dt}=g\sin\theta(1-\frac{V_y}{\sqrt{V_x^2+V_y^2}})$

We can eliminate the time dependence to get a differential equation for only $V_x$ and $V_y$ like this:

$\frac{dV_y/dt}{dV_x/dt}=\frac{dV_y}{dV_x}=\frac{1}{V_x}\big (V_y-\sqrt{V_x^2+V_y^2}\big )$

A little thought will show this is a homogenous differential equation, and so it can be separated if we make the substitution $u=\frac{V_y}{V_x}$ and $dV_y=udV_x+V_xdu$

When you separate the equation you will get $\frac{dV_x}{V_x}+\frac{du}{\sqrt{1+u^2}}=0$

Integrating this gives $\ln|V_x|+\ln|u+\sqrt{u^2+1}|=C_0=\ln|V_y+\sqrt{V_x^2+V_y^2}|$

Which means that $V_y+\sqrt{V_x^2+V_y^2}=C$

We use the initial condition $V_y(V_x=V_0)=0$ to find the constant is $C=V_0$

Now we use the condition that the terminal velocity will be when $V_x=0$ to find $V_{terminal}=V_y(V_x=0)=0.5V_0$

Let $x$ be the direction down the plane. At any moment, the acceleration in $x$ direction is the same as the retardation in tangential direction. So the magnitude of change in velocity in $x$ direction is the same as magnitude of change in velocity in tangential direction(which is simply the change in net velocity). Since the terminal velocity will be in $x$ direction, $v_x - 0 = v_0 - v_x \implies v_x = v_{\text{terminal}} = {v_0 \over 2 }$