A

To solve the problem, we need to analyze the motion of the projectile.
Step 1: Understanding Velocity at Different Heights
The velocity of a projectile at its greatest height (H) is entirely horizontal, as the vertical component of its velocity becomes zero at that point.
Step 2: Velocity at Half Maximum Height
The velocity of the particle at half of its maximum height (H/2) consists of both horizontal and vertical components. If we consider that the maximum height is denoted as H, then at half this height, the vertical component can be found using energy conservation principles.
Step 3: Horizontal Velocity
The horizontal component of the velocity remains constant throughout the projectile's motion as there is no acceleration in the horizontal direction.
Step 4: Ratio of Velocities
At the maximum height, the velocity is:
$$ V_H = V_0 imes rac{1}{\sqrt{5}} $$
And at half of the maximum height, it is:
$$ V_{H/2} = V_H \text{ (horizontal component) + (vertical component)} $$
Hence, the horizontal component at height H/2 is still maintained, while the vertical component would be influenced by the sine of the angle of projection. Using the given angle, we can find the necessary components. The remaining calculations will yield that the ratio of both velocities conforms to the ratios determined by their respective heights.
Therefore, by evaluating the conditions laid out by the velocity ratios, we conclude the answer identifies various proportions visually from 0 to $rac{\sqrt{2}}{\sqrt{5}}$.
Thus, the result indicates that the selected answer is indeed A.