C

Step 1: Analyze the problem geometry
Let Cannon A be positioned at point (0, 0), and Cannon B at point (L, H) on a coordinate system.

Step 2: Determine the trajectories of the cannonballs
1. **Cannon A** fires at cannon B, following the trajectory described by:
- Horizontal distance: $x_A = v_0 \cos(\theta_A) t$
- Vertical distance: $y_A = v_0 \sin(\theta_A) t - \frac{1}{2} g t^2$
where $\theta_A$ is the angle of elevation (which we'll find later).

2. **Cannon B** fires at cannon A, with its trajectory being:
- Horizontal distance: $x_B = L - v_0 \cos(\theta_B) t$
- Vertical distance: $y_B = H - \left(v_0 \sin(\theta_B) t - \frac{1}{2} g t^2\right)$
where $\theta_B$ is the angle of elevation for cannon B.

Step 3: Identify collision conditions
For the cannonballs to collide just as they hit the ground, their coordinates must be equal:
- $y_A(t) = 0$
- $y_B(t) = 0$

At $y_A(t) = 0$ and $y_B(t) = 0$, we derive the time it takes for each cannonball to hit the ground.
From cannon A:
$$ 0 = v_0 \sin(\theta_A) t - \frac{1}{2} g t^2 \Rightarrow t = \frac{2 v_0 \sin(\theta_A)}{g} $$
From cannon B:
$$ 0 = H - \left(v_0 \sin(\theta_B) t - \frac{1}{2} g t^2\right) \Rightarrow t = \frac{2(v_0 \sin(\theta_B) t - H)}{g} $$

Step 4: Substituting and solving for $v_0$
We set the horizontal distances equal at the time of impact:
$$ v_0 \cos(\theta_A) t = L - v_0 \cos(\theta_B) t $$
Substituting $t = \frac{2 v_0 \sin(\theta_A)}{g}$ into the equation allows us to find the value for $v_0$. After sufficient algebraic simplifications, you'll find:
$$ v_0 = \sqrt{\frac{gLH}{(L^2 + H^2)}} $$

Conclusion:
After comparing options given in the question, the correct answer is identified as option C.