How do I find the angle with respect to the ground that will cause a projectile launched from the origin to land at the origin? Given an arbitrary non-zero constant horizontal acceleration (like a breeze perhaps), how would one find the angle with respect to the ground (x-axis) that will cause a projectile launched from the origin to land at the origin?
 A: Note that a constant horizontal acceleration is not a realistic model for the force due to a breeze, but putting that aside: the basic kinematics equations for 2D motion of an object undergoing acceleration are the following:
$$\Delta x = v_{0x} t + \frac{1}{2} a_x t^2$$
$$\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$$
Where $\Delta x$ and $\Delta y$ are the changes in position in the x and y directions, respectively, $v_{0x}$ and $v_{0y}$ are the initial speeds in the x and y directions, respectively, and $a_x$ and $a_y$ are the accelerations in the x and y directions, respectively.  We can substitute for the speeds using the overall initial speed $v_0$, and the launch angle $\theta$ with
$$v_{0x} = v_0 cos(\theta)$$
$$v_{0y} = y_0 sin(\theta)$$
When you say "a projectile is launched from the origin and lands at the origin", you're saying that $\Delta x = \Delta y = 0$ for the same $t$.
$$0 = v_0 cos(\theta) t + \frac{1}{2} a_x t^2$$
$$0 = v_0 sin(\theta) t + \frac{1}{2} a_y t^2$$
If we divide both equations by $t$, we simplify the situation (while discarding the obvious solution in which the projectile goes nowhere), we get
$$0 = v_0 cos(\theta) + \frac{1}{2} a_x t$$
$$0 = v_0 sin(\theta) + \frac{1}{2} a_y t$$
Solving both equations for $t$,
$$t = - \frac{2 v_0 cos(\theta)}{a_x}$$
$$t = - \frac{2 v_0 sin(\theta)}{a_y}$$
Since the projectile must have $\Delta x = \Delta y = 0$ at the same time, we can let each $t$ be equal to each other, 
$$- \frac{2 v_0 cos(\theta)}{a_x} = - \frac{2 v_0 sin(\theta)}{a_y}$$
Solving for $\theta$:
$$tan(\theta) = \frac{a_y}{a_x}$$
$$\theta = tan^{-1}(\frac{a_y}{a_x})$$
Which gives us the necessary launch angle.
