A ball of mass $m$ is launched into the air. Ignore air resistance, but assume that there is a wind that exerts a constant force $F_0$ in the -$x$ direction. In terms of $F_0$ and the acceleration due to gravity $g$, at what angle above the positive $x$-axis must the ball be launched in order to come back to the point from which it was launched?

$ \textbf{(A)}\ \tan^{-1}(F_0/mg)$

$\textbf{(B)}\ \tan^{-1}(mg/F_0)$

$\textbf{(C)}\ \sin^{-1}(F_0/mg)$

$\textbf{(D)}\ \text{the angle depends on the launch speed}$

$\textbf{(E)}\ \text{no such angle is possible}$
1. **Identify the forces and accelerations:**
   - The ball experiences a constant force \( F_0 \) in the \(-x\) direction.
   - The ball also experiences a gravitational force \( mg \) in the \(-y\) direction.
   - Therefore, the accelerations are:
     \[
     a_x = -\frac{F_0}{m}, \quad a_y = -g
     \]

2. **Determine the time of flight:**
   - The vertical component of the initial velocity is \( v_{0y} \).
   - The time to reach the maximum height is \( t = \frac{v_{0y}}{g} \).
   - The total time of flight (up and down) is:
     \[
     t_{\text{total}} = 2 \cdot \frac{v_{0y}}{g} = \frac{2v_{0y}}{g}
     \]

3. **Determine the horizontal displacement:**
   - The horizontal component of the initial velocity is \( v_{0x} \).
   - The horizontal acceleration is \( a_x = -\frac{F_0}{m} \).
   - The horizontal displacement \( x(t) \) is given by:
     \[
     x(t) = v_{0x} t + \frac{1}{2} a_x t^2
     \]
   - For the ball to return to its starting point, \( x(t_{\text{total}}) = 0 \):
     \[
     0 = v_{0x} \left( \frac{2v_{0y}}{g} \right) + \frac{1}{2} \left( -\frac{F_0}{m} \right) \left( \frac{2v_{0y}}{g} \right)^2
     \]

4. **Simplify the equation:**
   - Substitute \( t_{\text{total}} = \frac{2v_{0y}}{g} \):
     \[
     0 = \frac{2v_{0x} v_{0y}}{g} - \frac{F_0}{m} \cdot \frac{2v_{0y}^2}{g^2}
     \]
   - Factor out \( \frac{2v_{0y}}{g} \):
     \[
     0 = \frac{2v_{0y}}{g} \left( v_{0x} - \frac{F_0 v_{0y}}{mg} \right)
     \]
   - Since \( \frac{2v_{0y}}{g} \neq 0 \), we have:
     \[
     v_{0x} = \frac{F_0 v_{0y}}{mg}
     \]

5. **Relate \( v_{0x} \) and \( v_{0y} \) using the launch angle \( \theta \):**
   - By definition, \( \tan{\theta} = \frac{v_{0y}}{v_{0x}} \):
     \[
     \tan{\theta} = \frac{v_{0y}}{\frac{F_0 v_{0y}}{mg}} = \frac{mg}{F_0}
     \]
   - Therefore, the launch angle \( \theta \) is:
     \[
     \theta = \tan^{-1} \left( \frac{mg}{F_0} \right)
     \]

Conclusion:
\[
\boxed{\theta = \tan^{-1} \left( \frac{mg}{F_0} \right)}
\]