# Trajectory function of projectile motion

1. Nov 19, 2012

### Arroway

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:
$$x(t)=\frac{cp_0cos\theta}{F}ln\left \{ \frac{\sqrt{E_0^2+c^2P^2(t)-c^2p_0^2sen^2\theta}+cP(t)}{E_0-cp_0sen\theta} \right \}\\\\ y(t)=\frac{1}{F}\left \{ E_0-\sqrt{E_0^2+c^2P^2(t)-c^2p_0^2sen^2\theta} \right \}$$
For which he found the following function:
$$y(x)=\frac{E_0}{F}-\frac{E_0}{F}cosh\left [ \frac{Fx}{p_occos\theta} \right ]+\frac{p_0csen\theta}{F}senh\left [ \frac{Fx}{p_0ccos\theta} \right ]$$
I'm having some trouble with this calculation because of that $cP(t)$ term. I've tried backtracking as well, but it didn't work. I'm feeling stupid. :(

2. Nov 19, 2012

### Bill_K

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.

3. Nov 20, 2012

### Arroway

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.
$P(t)=Ft-p_0sin\theta$ (I think the author did that to simplify the equations) and I'm assuming $F$ is the weight.

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

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