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 $\vec{r_0}$.
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...
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:
$\phi \rightarrow \theta; |\vec{v_0}| \rightarrow \infty$
$\phi \rightarrow \dfrac{\pi}{2}; |\vec{v_0}| \rightarrow \infty$
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?
We are given $h$ and $d$ and we choose an arbitary launch angle $\phi$ between $\theta$ and $\frac\pi2$.
In the plane of the projectile's trajectory, let $v_x$ be the horizontal component of the velocity vector. Let $t$ be the time spent in reaching $r_f$ from $r_0$. One of the equations of motion tells us that for an object in constant acceleration in one dimension like this projectile, $$u=u_0+v_0t+\frac12at^2$$ where $u$ stands for displacement. (The $v_0$ here is a scalar, as compared to the vector $\vec{v_0}$.)
Now consider the projectile's motion in the vertical axis. $u_0$ is obviously 0 (taking $r_0$ as the origin), $u=h$ and $a=-g$; $v_0$ is seen to be $v_x\tan\phi$. Yet we also know that the projectile's horizontal velocity is constant, so we have $t=\frac d{v_x}$. Substituting all this into the equation of motion we have $$h=(v_x\tan\phi)\frac d{v_x}-\frac12g\left(\frac d{v_x}\right)^2$$ $$h=d\tan\phi-\frac12g\frac{d^2}{v_x^2}$$ $$d\tan\phi-h=\frac12g\frac{d^2}{v_x^2}$$ $$\frac2g(d\tan\phi-h)=\frac{d^2}{v_x^2}$$ $$v_x^2=\frac{d^2}{\frac2g(d\tan\phi-h)}=\frac{gd^2}{2(d\tan\phi-h)}$$ $$v_x=\sqrt{\frac{gd^2}{2(d\tan\phi-h)}}=d\sqrt{\frac g{2(d\tan\phi-h)}}$$ The velocity vector in the trajectory's plane is $(v_x,v_0)=(v_x,v_x\tan\phi)$. The speed $|\vec{v_0}|$ is then $$\sqrt{v_x^2+(v_x\tan\phi)^2}=v_x\sqrt{1+\tan^2\phi}=v_x\sec\phi$$ $$=d\sec\phi\sqrt{\frac g{2(d\tan\phi-h)}}$$