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Trajectory function of projectile motion

  1. Nov 19, 2012 #1
    I've been studying this rather interesting article about projectile motion in special relativity. The thing is, I can't understand how the author found the trajectory function. He says that he did it by solving the following parametric equation:
    [tex]x(t)=\frac{cp_0cos\theta}{F}ln\left \{ \frac{\sqrt{E_0^2+c^2P^2(t)-c^2p_0^2sen^2\theta}+cP(t)}{E_0-cp_0sen\theta} \right \}\\\\
    y(t)=\frac{1}{F}\left \{ E_0-\sqrt{E_0^2+c^2P^2(t)-c^2p_0^2sen^2\theta} \right \}[/tex]
    For which he found the following function:
    [tex]y(x)=\frac{E_0}{F}-\frac{E_0}{F}cosh\left [ \frac{Fx}{p_occos\theta} \right ]+\frac{p_0csen\theta}{F}senh\left [ \frac{Fx}{p_0ccos\theta} \right ][/tex]
    I'm having some trouble with this calculation because of that [itex]cP(t)[/itex] term. I've tried backtracking as well, but it didn't work. I'm feeling stupid. :(
  2. jcsd
  3. Nov 19, 2012 #2


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    Don't know what any of this stands for, and the paper is behind a paywall. What is F, what is P(t)?? (I assume that sen, senh mean sin, sinh.)

    Let X = Fx/(p0c cosθ), let Q = stuff inside the √, and drop the 0 subscripts on p0, E0.

    eX = (√Q + cP)/(E - cp sinθ) = (E + cp sinθ)(√Q + cP)/(E2 - c2p2sin2θ)
    e-X = ... = (E - cp sinθ)(√Q - cP)/(E2 - c2p2sin2θ)

    cosh X = (E√Q + c2pP sin2θ)/(E2 - c2p2sin2θ)

    sinh X = (cp sinθ√Q + cPE)/(E2 - c2p2sin2θ)

    From which, -E cosh X + pc sinθ sinh X = -√Q, which is (almost) y.
  4. Nov 20, 2012 #3
    I'm sorry, I didn't write what any of these terms meant because my issue was with the algebra, not the problem's physics.
    [itex]P(t)=Ft-p_0sin\theta[/itex] (I think the author did that to simplify the equations) and I'm assuming [itex]F[/itex] is the weight.

    Thank you for the help, it didn't occur to me that if I multiplied [itex]e^{-X}[/itex] by that I would get rid of the [itex]Q[/itex] in the denominator. Like I said, I'm stupid. :(
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