C

Step 1: In projectile motion, the following observations can be made for a ball projected at an initial velocity of .

Velocity at half the maximum height (A-I):
At half the maximum height, the vertical component of the velocity is reduced to half of its initial vertical component, while the horizontal component remains the same. Therefore, the resultant velocity is given by:
$$ v_{half} = rac{v imes sin( heta)}{2} + v imes cos( heta). $$
This matches with (I).

Velocity at the maximum height (B-II):
At maximum height, the vertical component of the velocity is zero, while the horizontal component remains. Therefore, the magnitude of velocity at the maximum height equals the initial horizontal component:
$$ v_{max height} = v imes cos( heta). $$
This matches with (II).

Change in velocity when it returns to the ground (C-III):
The change in velocity when the ball returns to the ground is equal to twice the initial vertical component of the projection, as the velocity will have a direction that is opposite upon returning to the ground:
$$ ext{Change in velocity} = 2 imes v imes sin( heta). $$
This matches with (III).

Average velocity when it reaches the maximum height (D-IV):
The average velocity up to the maximum height is the initial velocity in the direction of motion divided by 2:
$$ V_{average} = rac{v imes sin( heta)}{2} $$ - this corresponds to (IV).

Therefore, the correct pairings are:
A-I, B-II, C-III, D-IV.
Hence, the answer is Option C.