# Obtaining a general equation for velocity (in 2D projectile motion) [closed]

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?

• The equation you've been given is just a trajectory on an $(x,y)$ graph and contains no information about the time dependence of the particle's position. Without any extra information you cannot calculate the velocity because you don't know the particle position as a function of time. – John Rennie Jun 12 '17 at 6:27
• @John Darn...totally failed to see that. Thanks! (Do I have to delete the question now?) – Alan Jun 12 '17 at 6:31
• Well, there could be an answer. For example the obvious extra information is that $d^2y/dt^2 = g$, and including this allows you to get the velocity. It's up to you if you want to let the question stand and see if anyone answers it. – John Rennie Jun 12 '17 at 6:33

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.

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, $$v_s=v_s(initial)-gt$$

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:

$$y = x \tan \theta - \frac{ g x^2}{2 (u \cos \theta)^2}$$

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.