C

Step 1: Identify the components of the trajectory equation. The trajectory is given as:
$$y = x - \frac{g}{2v^2} x^2$$
where g = 10 m/s².

Step 2: The angle of projection ($\theta$) can be determined from the trajectory equation. The equation represents a parabolic motion, where the term for gravity affects how the y-value changes with x. The angle can be derived from the slope at the origin of the parabola – which, if equal to the change in y per change in x, gives you:
$$tan(\theta) = \frac{1}{m}$$ (slope of the initial trajectory). Since m = 2, therefore, the angle is:
$$\theta = tan^{-1}(1/2)$$
which corresponds to option D.

Step 3: Now calculate the angle of velocity after 4 seconds for projectile motion, which involves computing the vertical velocity at that point:
$v_{y} = v_{0y} - gt$ where $g = 10 m/s^2$.
If you visualize this, you would see that at 4 seconds, the angle would need additional calculations based on v0. This needs deeper analysis to find the velocity slope after 4 seconds. Without specific values for v0, we can't numerically compute beyond this point.

Step 4: Maximum height and trajectory aspects need careful assessments of peak conditions (when vertical velocity becomes zero). Correlate back to horizontal displacement as a function of height for key determinations.

Hence, option C is found to represent the angle of projection. Other aspects require more detailed velocity calculations to confirm their respective alignments.