# Getting pairs of angle and velocity for a projectile to a given destination

I'm trying to calculate the initial velocity $v_0$ and angle $\theta$ for a given destination $(x, y)$ with a launch height of $y_0$. Obviously there will be a set of pairs of velocity and angle that will pass through the destination point. This set is given by $$\left\{(v_0,\theta) \middle|y = - \frac{g}{2 v_0^2 \cdot \cos^2(\theta)} \cdot x^2 + \tan(\theta) \cdot x +y_0 \right\} \ .$$

I have already looked at this questions:

Solving for initial velocity required to launch a projectile to a given destination at a different height

How to get the angle needed for a projectile to pass through a given point for trajectory plotting

So for example I could rewrite the set to $$\left\{ (v_0,\theta) \middle| v_0 = \frac{1}{\cos(\theta)}\sqrt{\frac{\frac{1}{2} g x^2}{x \tan(\theta)+y_0}} \right\}\ .$$

Is there an easy way to compute this set for one given destination without iterating over all possible velocities or angles?

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There is more than one solution. Choose an angle, arbitrarily if you like, and compute the velocity you need. So I don't understand what you mean by "iterating over all possible velocities or angles". – Michael Brown Mar 12 '13 at 0:24
what are the other information provided related to the coordinate $(x,y)$ – Dimensionless Mar 12 '13 at 2:34
I want to calculate not only one pair of angle and velocity, I want to compute the set of pairs. For me the only way to proximate this set is to iterate over all angles and calculate the velocity or the other way around. @Akash I don't know any thing else about the destination. In fact I know launch point and the destination. – Sam Mar 12 '13 at 10:39
then you could use this equation for the trajectory of the projectile in term's of &R& $y$ = $x(1-\frac{x}{R})tan(\alpha)$ where $\alpha$ is the angle of projection – Dimensionless Mar 12 '13 at 14:26
@Akash I have realy no idea want you mean. What is $R$, is it the range? What is the difference between $\theta$ and $\alpha$? – Sam Mar 12 '13 at 15:36