A

Step 1: Identify the variables involved. We need to find the initial speed () required for the ball to clear the wall of height 'h' at a horizontal distance 'x'.

Step 2: Use the equations of projectile motion. The range R of a projectile is given by:
R = \frac{u^2 \sin(2\theta)}{g}, where 'u' is the initial speed, 'g' is the acceleration due to gravity, and 'theta' is the angle of projection.
Here, the horizontal distance (R) is equal to 'x'.

Step 3: The height 'h' achieved by the projectile at the horizontal distance 'x' is given by:
h = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)}.

Step 4: To find the minimum speed required to clear the wall, we need to maximize the range for a given maximum height. This can be equivalent to setting up a right triangle where the vertical and horizontal components need to just clear the wall. After manipulating the projectile motion equations, we can derive:
u = \sqrt{\frac{g h}{2(1 - \frac{h}{h_{max}})}}
For the case of the minimum speed, substituting the conditions yields:

Final Step: The minimum speed, u_min = \sqrt{\frac{g h}{x}}.

Thus, we conclude that the minimum required speed to just clear the wall is derived from the parameters provided.