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My ultimate goal is to find the angle of elevation necessary to launch a projectile from the origin to (x,y) with initial velocity V and under gravitational acceleration g. Wind resistance is ignored.

This is where I've started, and the path I've taken. First, I took the parametrics for x and y.

x = V * cosθ * t

y = V * sinθ * t - (g / 2) * t^2

-Then, I found t in terms of x, V and θ...

t = x / (V * cosθ)

-I plugged that value in for t in the y equation...

y = V * sinθ* (x / (V * cosθ)) - (g / 2) * (x^2 / (V^2 * (cosθ)))

y = x * tan⁡θ - (g * x^2) / (2 * V^2 * (cos⁡θ)^2)

I don't know what to do from here. I need to find θ in terms of x, y, V, and g, but it seems impossible to get all of the θ's into a single term. (x, y, V and g will be known when I use the equation).

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The second part of the expression for $y$ should have $\cos^2\theta$ at the bottom, not $\sin\theta$. What is the question? In general there may be a number of combinations of $V$, $\theta$ that get the curve to pass through a specific point. – André Nicolas Nov 17 '11 at 23:31
thanks, fixed that error, the one that's up now is what I have. The question issue is fixed (see end of post) – Miles Rufat-Latre Nov 17 '11 at 23:35
up vote 3 down vote accepted

Our equation looks like $y=x\tan\theta+\dfrac{k}{\cos^2\theta}x^2$ for a certain constant $k$.

Rewrite this as $y=x\tan\theta+kx^2\,\sec^2\theta$, and use the fact that $\sec^2\theta=1+\tan^2\theta$.

We end up with a quadratic equation in the variable $w=\tan\theta$. Solve for $w$, using the Quadratic Formula. Now we have an expression for $\tan\theta$. Apply the $\arctan$ function to get $\theta$.

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I'll try that, come back and tell you if I got it. Sounds like the right solution. – Miles Rufat-Latre Nov 17 '11 at 23:56
That worked! Thanks! – Miles Rufat-Latre Nov 18 '11 at 4:07
@Miles Rufat-Latre: Thanks for letting me know. I was worried that a coding detail would make you think it didn't. – André Nicolas Nov 18 '11 at 5:59
lol, Wolfram Alpha times out trying to solve it – Miles Rufat-Latre Nov 19 '11 at 1:40
@Miles Rufat-Latre: If you wish, I can solve a specific numerical example. I would have to know all the numbers, $g$, initial speed, $x$ and $y$. Now as to why Wolfram Alpha misbehaves, many combinations of the parameters yield no answer. If you throw a javelin with the very slow initial speed of $1$ meter per second, it will never reach the point $(1000,20)$. – André Nicolas Nov 19 '11 at 2:01

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