Position Vector for a Projectile I'm new here. Any advice is greatly appreciated.
$$r(t) = (v_0\cos \theta)t\mathbf i + [h+(v_0\sin \theta)t- \frac 12gt^2]\mathbf j\\$$
I'm trying to find a combination of $\theta$ and $v_0$ that yields a maximum height and an $x$-intercept that is given. Assume we know $h$ and $g$. 
For example, suppose we're asked to find a vector valued position function, using the position vector for a projectile. The vector valued position function must reach a maximum height of 64 blah units and have an $x$-intercept at $x=$ 7 blah units. Assume $h=$ 24 and $g=$ 32. Any guidance is appreciated.
 A: You have a few equations that you need to satisfy.
$$\begin{align}
v_0t_1\cos\theta&=x_{intercept}&(1)\\
h+v_0t_1\sin\theta-\frac12gt_1^2&=0&(2)\\
h+v_0t_2\sin\theta-\frac12gt_2^2&=h_{max}&(3)\\
t_2&=\frac{v_0\sin\theta}g&(4)
\end{align}$$
You have four main variables: $v_0, t_1, t_2, \theta$. $t_1$ is when the object hits the ground. $t_2$ is when the object reaches its maximum height. The first two equations deal with the fact that at $t=t_1$, the object will be at position $x_{intercept}\boldsymbol i+0\pmb j$. The third and fourth equations are used used to find the time at which the projectile is at its maximum height.
Substitution will get you the answers. Substituting $(4)$ into $(3)$ yields
$$\begin{align}
&v_0^2\sin^2\theta-\frac12(v_0^2\sin^2\theta)=g(h_{max}-h)\\
&\implies v_0\sin\theta=\sqrt{2g(h_{max}-h)}
\end{align}$$
Now, you can substitute that into both$(2)$ and $(3)$ to make both equations only depend on $t$, which you can solve using the quadratic method. Once you solve for both $t_1$ and $t_2$, you can rewrite $(1)$ and $(4)$ as 
$$\begin{align}
v_0\cos\theta&=\frac{x_{intercept}}{t_1}\\
v_0\sin\theta&=gt_2&\implies\\
v_0^2\sin^2\theta+v_0^2\cos^2\theta=v_0^2&=\frac{x_{intercept}^2}{t_1^2}+g^2t_2^2&\implies\\
v_0&=\sqrt{\frac{x_{intercept}^2}{t_1^2}+g^2t_2^2}
\end{align}$$
The right-hand side is a constant since you solved for $t_1$ and $t_2$. Finally, use $(1)$ or $(4)$ (unless you hate yourself and want to use $(2)$ or $(3)$) to solve for $\theta$
$$\theta=\arcsin\left(\frac{gt_2}{v_0}\right)$$
