Calculating angle for trajectory with exit position not at 0,0

[xal]

I'd like to calculate the angle to shoot a projectile from a cannon in order to hit a target. However, the position that the projectile will be exiting from will not be at 0, 0 -- it will vary according to the angle of the cannon itself. The projectile will be exiting from the tip of the cannon, but the cannon will be rotated about a different point (both points, as well as the target position, will be on the same plane though).

If the projectile was exiting from 0, 0, then I could use this:
$$\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\mbox {{\tt `\&+-`}<br /> } \left( \sqrt {{{\it v0}}^{4}-{g}^{2}{x}^{2}-2\,gy{{\it v0}}^{2}}<br /> \right) }{gx}}$$

However, in this case, I was considering replacing $$x$$ and $$y$$ with $$x-l\cos \left( \theta \right)$$ and $$y-l\sin \left( \theta \right)$$ respectively, where $$l$$ would be the distance away the tip.

That results in this:
$$\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\sqrt {{{\it v0}}^{<br /> 4}-g \left( g \left( x-l\cos \left( \theta \right) \right) ^{2}+2\,<br /> \left( y-l\sin \left( \theta \right) \right) {{\it v0}}^{2} \right) <br /> }}{g \left( x-l\cos \left( \theta \right) \right) }}<br /> <br />$$

Different form:
$$-1/2\,{\frac {g \left( x-l\cos \left( \theta \right) \right) ^{2}<br /> \left( \tan \left( \theta \right) \right) ^{2}}{{v}^{2}}}+x\tan<br /> \left( \theta \right) -1/2\,{\frac {g \left( x-l\cos \left( \theta<br /> \right) \right) ^{2}}{{v}^{2}}}-y+l\sin \left( \theta \right) =0$$

But I am not sure if this is a way to do it, and I am not sure to go about solving for the angle either.

 

[member 553083]

Any help would be greatly appreciated! The above equation can be solved for theta by using the Quadratic Formula. Specifically, you can rearrange the equation to the form ax^2 + bx + c = 0 and then solve for x. Once you have x, you can take its inverse tangent to find the angle theta.