A

Step 1: Identify the range formula for projectile motion given by \( R = \frac{v^2 \sin(2\theta)}{g} \).
Step 2: Calculate the ranges for both projectiles:
Projectile 1 (\( \theta_1 = 30^\circ \)):
\[ R_1 = \frac{v^2 \sin(60^\circ)}{g} = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{v^2 \sqrt{3}}{2g} \]
Projectile 2 (\( \theta_2 = 60^\circ \)):
\[ R_2 = \frac{v^2 \sin(120^\circ)}{g} = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{v^2 \sqrt{3}}{2g} \]
Both have the same range: \( R_1 = R_2 \).

Step 3: Find maximum heights using the formula \( H = \frac{v^2 \sin^2(\theta)}{2g} \).
Projectile 1:
\[ H_1 = \frac{v^2 \sin^2(30\circ)}{2g} = \frac{v^2 \cdot \left(\frac{1}{2}\right)^2}{2g} = \frac{v^2}{8g} \]
Projectile 2:
\[ H_2 = \frac{v^2 \sin^2(60\circ)}{2g} = \frac{v^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3v^2}{8g} \]
Thus, \( H_1 : H_2 = 1 : 3 \).

Step 4: Determine time of flights using the formula \( T = \frac{2v \sin(\theta)}{g} \).
Projectile 1:
\[ T_1 = \frac{2v \cdot \frac{1}{2}}{g} = \frac{v}{g} \]
Projectile 2:
\[ T_2 = \frac{2v \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}v}{g} \]
Thus, \( T_1 : T_2 = 1 : \sqrt{3} \).

Conclusion: Aligning ranges, maximum heights, and times of flights, we complete the answer as per the correlation specified in the question.