A

Step 1: Identify the time of flight for each projectile. The time of flight for a projectile is given by the formula:
$$ T = \frac{2u \sin\theta}{g} $$
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
For the first projectile (angle 60°):
$$ T_1 = \frac{2u \sin(60°)}{g} = \frac{2u (\frac{\sqrt{3}}{2})}{g} = \frac{u \sqrt{3}}{g} $$
For the second projectile (angle 30°):
$$ T_2 = \frac{2u \sin(30°)}{g} = \frac{2u (\frac{1}{2})}{g} = \frac{u}{g} $$
Since the angles are different, the values of \( T_1 \) and \( T_2 \) will be different.

Step 2: Now, calculate the range of each projectile. The range is given by the formula:
$$ R = \frac{u^2 \sin(2\theta)}{g} $$
For 60°:
$$ R_1 = \frac{u^2 \sin(120°)}{g} \quad (\sin(120°) = \frac{\sqrt{3}}{2})$$
For 30°:
$$ R_2 = \frac{u^2 \sin(60°)}{g} \quad (\sin(60°) = \frac{\sqrt{3}}{2})$$
Thus, ranges will also differ as the angles differ.

Step 3: Maximum height acquired is given by:
$$ H = \frac{u^2 \sin^2\theta}{2g} $$
For 60°:
$$ H_1 = \frac{u^2 (\sin(60°))^2}{2g} = \frac{u^2 (\frac{3}{4})}{2g} $$
For 30°:
$$ H_2 = \frac{u^2 (\sin(30°))^2}{2g} = \frac{u^2 (\frac{1}{4})}{2g} $$
Again, max heights will be different.

Conclusion: Only the time of flight remains the same when considering the individual effects of the angles. Hence, the correct option is Time of Flight (Option A).