while studying some material on ballistic trajectories (the basic gravity-only parabola), I've tried to come up with closed-form expressions for another case, where in addition to gravity we add a constant force on the x-direction:


With a first method I obtained the equations of motion: $$ \left\{ \begin{array}{c} x(t)=-\frac{\beta}{2}t^2 +(v_0\cos\alpha)t;\ \text{with $\beta=f/m$} \\ y(t)=-\frac{g}{2}t^2 +(v_0\sin\alpha)t\ \\ \end{array} \right. $$

But the problem was that I couldn't find an analytical form for $y=y(x)$

A second method from: $\ddot y=-g, \ \ddot x=-\beta $

Which gives: $\ddot y=\frac{g}{\beta}\ddot x$ , then by integrating: $\dot y -v_{0,y}=\frac{g}{\beta}\dot x -\frac{g}{\beta}v_{0,y}$

Until you get: $y=\frac{g}{\beta}x+t(v_{0,y}-\frac{g}{\beta}v_{0,x})$

$\ddot x=-\beta $ gives $\dot x=v_{0,x}+\beta t$ , therefore $t=\frac{\dot x -v_{0,x}}{-\beta}$ Again I don't know how to follow on as I get this differential equation: $$y=\frac{g}{\beta}x+\frac{v_{0,y}-\frac{g}{\beta}v_{0,x}}{v_{0,x}-\dot x}$$

The "homogeneous" (without y) version of this equation hints at the Lambert W function, which clearly shouldn't belong in this ballistics problem (so this second attempt is also faulty). Is there any clear analytical help?


Is there really a solution $y=y(x)$ to your problem? Part of me thinks not.

Another constant force in the x-direction is like having gravity in a different, non-downward direction. One method to solve this is to take your solution for downward gravity and rotate it in the x-y plane until the old downward direction matches your new gravity direction. The problem with this is that for a general rotation of a parabola, I can't see there being a solution of the form $y=y(x)$. Try sketching a rotated parabola; you'll find the picture doesn't pass the vertical line test for functions $y=f(x)$. So the best you could do is a more general solution like $f(x,y)=\text{const}$.

  • $\begingroup$ You're right, and plotting for these kinds of parametric equations says so (skewed parabola). But isn't a friction force $\vec f=-b\vec v$ a textbook problem with closed solutions (and thus my problem above a specific case of it)? $\endgroup$ – N.E. Jan 26 '14 at 16:15
  • $\begingroup$ or is this vectorial form unsolvable as well? $\endgroup$ – N.E. Jan 27 '14 at 17:28
  • $\begingroup$ One can obtain solutions like $x(t)$ and $y(t)$. You can even plot the trajectory if you have the appropriate software. In this sense, it's a doable problem. But I don't think you can find a neat mathematical form like $y=f(x)$ for the above reason. One should be able to find, however, some other form like $f(x,y)=\text{const}$, though I haven't done it myself. $\endgroup$ – BMS Jan 27 '14 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.