A

To solve this problem, we will use the principles of projectile motion.
Step 1: Define the variables. Let \( h \) be the height of the plane, \( V \) be the horizontal speed of the plane, and \( u \) be the muzzle velocity of the shell at an angle \( \theta \) with the horizontal.
Step 2: The time \( t \) it takes for the shell to rise to height \( h \) can be found from the vertical motion equation:
\[ h = u \sin(\theta) t - \frac{1}{2} g t^2 \]
where \( g \) is acceleration due to gravity.
Step 3: The horizontal distance traveled by the shell when it reaches the height of the plane must equal the horizontal distance traveled by the plane in time \( t \):
\[ V t = u \cos(\theta) t \]
Step 4: We can eliminate \( t \) from these two equations to find an expression involving \( u \). From the horizontal equation, we have \( t = \frac{V}{u \cos(\theta)} \). Substitute \( t \) in the vertical motion equation:
\[ h = u \sin(\theta) \cdot \frac{V}{u \cos(\theta)} - \frac{1}{2} g \left( \frac{V}{u \cos(\theta)} \right)^2 \]
Simplifying gives us:
\[ h = V \tan(\theta) - \frac{g V^2}{2 u^2 \cos^2(\theta)} \]
Rearranging yields:
\[ \frac{g V^2}{2 u^2 \cos^2(\theta)} = V \tan(\theta) - h \]
Step 5: To find the minimum muzzle velocity, set \( an(\theta) = \frac{h}{V} \) at the point of maximum range.
Hence, using trigonometric identities and simplifying, we can derive that the minimum muzzle velocity \( u \) must satisfy:
\[ u = \sqrt{\frac{g h}{\sin(2\theta)}} \] where \( \theta \) is optimized for range.
This gives us the required expression in terms of the variables defined above.
Conclusion: The minimum muzzle velocity required to hit the plane can be expressed using the above analysis, confirming option A as correct.