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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?

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  • $\begingroup$ 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. $\endgroup$ – John Rennie Jun 12 '17 at 6:27
  • $\begingroup$ @John Darn...totally failed to see that. Thanks! (Do I have to delete the question now?) $\endgroup$ – Alan Jun 12 '17 at 6:31
  • $\begingroup$ 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. $\endgroup$ – John Rennie Jun 12 '17 at 6:33
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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.

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

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