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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:

diagram

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?

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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 $f(x,y)=\text{const}$.

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  • $\begingroup$ 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

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