Obtaining a general equation for velocity (in 2D projectile motion) I'm trying to obtain a general equation for the instantaneous velocity of a projectile moving on a Cartesian plane.
I began with the equation for a projectile's trajectory (air resistance neglected):
$$y = x(\tanθ) - \frac {gx^2}{(u^2)(\cosθ)^2}$$
where $u$ is the projection velocity, and $θ$ is the projection angle.
I then sought to differentiate the above-mentioned equation with respect to time. This yielded:
$$y' = x'(\tanθ) - \frac {2gxx'}{(u^2)(\cosθ)^2}$$
Where $'$ stands for a differential with respect to time.
Now, re-writing the equation:
$$v_y = v_x(\tanθ) - \frac {2gxv_x}{(u^2)(\cosθ)^2}$$
Where $v_y$ and $v_x$ are the $y$ and $x$ components of instantaneous velocity.
My issue?
I can't seem to be able to get the last equation in terms of the variables $v_y$ and $v_x$ alone (I can't seem to eliminate the $x$).
My question:

Is it possible to obtain a general equation for instantaneous velocity with  $v_y$ and $v_x$ as the only variables? If so, how do I go about it?

 A: Let's start out from scratch, assume the velocity components to be:
$v_x=v\cos\theta, v_y=v\sin\theta-gt$ 
(We know the $y$ acceleration is $g$, which is obvious from differentiating twice w.r.t $t$)
Which implies $x(t)=v\cos\theta t, y(t)=v\sin\theta t-0.5gt^2$ (starting from $(0,0)$).
Combining these two (i.e. getting rid of $t$), we get:
$$y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta}$$
This is the same as your equation, but with $v=\frac{u}{\sqrt2}$. 
This works, IMO.
A: The equation you've given is the equation of trajectory where the x and y coordinates of the position/displacement of projectile is given.
For finding $v_x$ , it remains constant in time( horizontal component remains unchanged with no air resistance)
For $v_s$ it is just given according to $v=u+at$
So, $$v_s=v_s(initial)-gt$$
A: You have an equation for the trajectory which is time-independent.
$$ y = x \tan \theta - \frac{ g x^2}{(u \cos \theta)^2} $$
By differentiating in terms of $x$ you get the slope of the trajectory
$$ \frac{{\rm d}y}{{\rm d}x} = \tan \theta - \frac{2 g x}{(u \cos \theta)^2} $$
Note that velocity vector is always tangent to the trajectory which means in terms of the velocity components
$$ \left. \frac{v_y}{v_x} = \frac{{\rm d}y}{{\rm d}x} \right\} v_y = \left(\tan \theta - \frac{2 g x}{(u \cos \theta)^2} \right) v_x $$
and since $v_x =u \cos \theta$ is constant, then
$$ v_y = u \sin \theta - \frac{2 g x}{u \cos \theta} $$
The above gives the vertical velocity component as a function of position $x$
If you want $v_y$ as a function of time then use  $x = t ( u \cos \theta)$ above, which comes out to be
$$ v_y = u \sin \theta - 2 g t $$
This is incorrect as the actual vertical speed should be $ v_y = u \sin \theta - g t $ which leads to believe you have made a mistake in the trajectory equation.
The actual trajectory should have been:
$$ y = x \tan \theta - \frac{ g x^2}{2 (u \cos \theta)^2}$$
A: Is your goal simply to write down an equation that relates $v_x$ and $v_y$, without reference to $x$, $y$, or $t$?
If so: Such an equation can’t exist, for the simple reason that $v_x$ is a constant while $v_y$ varies over the trajectory. If you found some relationship $f(v_x, v_y)=0$, you could differentiate it with respect to $t$:
$$
0 = \frac{d}{dt} f(v_x, v_y) = \frac{\partial f}{\partial v_x} \frac{d v_x}{dt} + \frac{\partial f}{\partial v_y} \frac{d v_y}{dt} = \frac{\partial f}{\partial v_y} \frac{d v_y}{dt}.
$$
From this, you could conclude that $v_y$ was a constant as well.
The only loophole in this argument is when the function $f$ is independent of $v_y$ ($\partial f/\partial v_y = 0$ above).  An equation of this form does exist: $v_x=v \cos \theta $.  It doesn’t contain $x$, $y$, or $t$,  but it doesn’t contain $v_y$ either.
