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Ok, if I am given an initial position (at the origin) and the final position coordinates $(x,y)$ as well as the initial velocity magnitude $V_0$. How do I find the correct initial angle that $V_0$ should be at?
I have the following work, for the horizontal part:
$x=V_0\cos(\theta)t $ and vertical motion:
$y=V_0\sin(\theta) t- \frac{1}{2}gt^2$ .

I don't know the time required for this trajectory so I solve for $t$ in the horizontal and plug into the vertical equation: What follows is an equation in terms of $\sin(\theta)$, $\cos(\theta)$ and $\cos^2(\theta)$.
As for the mathematics, I am trying to isolate the $\theta$ out of that equation (solve for it) but I can't. I've tried various trig identities but can't seem to do it. How should I proceed?

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  • $\begingroup$ in one word: numerically $\endgroup$ – Hagen von Eitzen Apr 25 '14 at 17:13
  • $\begingroup$ You should include the equation in terms of $\sin,\cos,$ and $\cos^2$ in the question. $\endgroup$ – Alexander Gruber Apr 25 '14 at 17:19
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To isolate take the g term to the LHS. Square both sides. Use sin^2 = 1-cos^2. this gives you a quadratic in cos^2 which you can solve.

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