Initial velocity to get from on known position to another?

[floyd0117]

I have a problem where I need to figure out the initial velocity vector \vec{v_0} of a projectile, in order for it to land at the final position \vec{r_f} = x_f\hat{x} + y_f\hat{y} + z_f\hat{z}, from initial position \vec{r_0}.
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The only knowns in the problem are \vec{r_0} and \vec{r_f}. Air resistance is neglected, so the the components of the net force on the projectile are

m\ddot{x} = 0

m\ddot{y} = 0

m\ddot{z} = -mg

So really we can choose any launch angle \phi, and find the necessary |\vec{v}|, or the other way around, to land us at \vec{r_f}. I think it sounds easier to choose a \phi and then find |\vec{v}|. So, I examine the limiting cases...
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Let's say d is the the distance between the initial and final positions in the x-y plane, that is;

d = |x_f\hat{x} + y_f\hat{y}|

and that h is the desired final height, h = z_f.

Then the angle \theta measured form the x-y plane to a line connecting (x_0, y_0 ,z_0) to (x_f, y_f, z_f) is smiply

\theta = \arctan{\dfrac{h}{d}}

So, our limiting cases are:

\phi \rightarrow \theta; |\vec{v_0}| \rightarrow \infty

\phi \rightarrow \dfrac{\pi}{2}; |\vec{v_0}| \rightarrow \infty
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So I can choose any angle between \dfrac{\pi}{2} and \theta, though angles close to those values will necessitate a very large initial velocity. My question is, how do I go from here, to determining |\vec{v_0}|? If I choose a \phi, how do I find a velocity that will get me to \vec{r_f}? It would seem that I need some function of v_0 in terms of both \phi (known, after choosing), and \vec{r_f}. Am I severely over thinknig this?

 

[haruspex]

how do I go from here

You don't go from there. Knowing the range of angles is not much help.
Just suppose the flight time is t and write an expression for where it will be at time t.

 

[A.T.]

My question is, how do I go from here, to determining |→v0||\vec{v_0}|? If I choose a ϕ\phi, how do I find a velocity that will get me to →rf\vec{r_f}?

The trajectory is a parabola. You know 2 points on it, and the slope (1st deviate) at one of them.