The bullets are fired at the same initial speed

To solve this problem, we can use basic principles of projectile motion and trigonometry. Let's break down the problem into smaller steps:

Step 1: Analyzing the problem
- The gun fires two bullets at angles of 60º and 30º horizontally.
- We need to determine the ratio of the maximum heights reached by the two bullets.

Step 2: Finding the initial velocities
- We know that the initial velocity of the bullets will have both horizontal and vertical components.
- The horizontal component remains the same for both bullets and is equal to the velocity of the bullets.
- Let's assume the initial velocity of the bullets is "v."
- The vertical component of the velocity for the 60º bullet is v * sin(60º) = v * √(3)/2.
- The vertical component of the velocity for the 30º bullet is v * sin(30º) = v * 1/2.

Step 3: Finding the time of flight
- The time of flight for a projectile can be determined using the formula T = 2 * (V0 * sin(θ)) / g, where V0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
- The time of flight for the 60º bullet is T1 = 2 * (v * √(3)/2) / g = √(3) * v / g.
- The time of flight for the 30º bullet is T2 = 2 * (v * 1/2) / g = v / g.

Step 4: Finding the maximum height
- The maximum height reached by a projectile can be determined using the formula H = (V0^2 * sin^2(θ)) / (2 * g), where V0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
- The maximum height reached by the 60º bullet is H1 = (v * √(3)/2)^2 / (2 * g) = 3v^2 / (8g).
- The maximum height reached by the 30º bullet is H2 = (v * 1/2)^2 / (2 * g) = v^2 / (8g).

Step 5: Finding the ratio of maximum heights
- The ratio of the maximum heights is H1 : H2 = (3v^2 / (8g)) : (v^2 / (8g)) = 3 : 1.

Therefore, the correct answer is option B) 3 : 1, as the ratio of the maximum heights reached by the two bullets is 3 : 1.