C

Step 1: The equation of the projectile motion is given as \(y = x - \frac{g x^2}{2 v_0^2 \cos^2\theta}\), where \(g = 10 \, m/s^2\) is the acceleration due to gravity. The path equations describe the trajectory of a projectile in terms of its horizontal and vertical displacement.

Step 2: To find the angle of projection \(\theta\), we can use the maximum height attained by the projectile, which occurs at \(\theta = 45º\) for maximum range on level ground. However, we still need to derive specific details from the projectile path equation.

Step 3: From the equation given, maximum height is determined when the vertical velocity is zero. At this position, using the formula \(h_{max} = \frac{v_0^2 \sin^2 \theta}{2g}\). This will help in verifying our initial assumptions.

Given the options and trajectory characteristics:
(i) The angle of projection correlates to achieving a maximum height of the projectile. Since it appears in the derived formula and generally for maximum distances, the angle that provides maximum range is usually \(45º\).

Therefore:
(iii) The maximum height attained corresponds with option C, which is \(45º\).
Therefore, the correct answer is C.