Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:

$$H = \frac{v^2 \sin^2 \theta}{2g}$$

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

$$H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 \times \frac{1}{4}}{2g} = \frac{v^2}{8g}$$

For the projectile projected at 60°, the maximum height is:

$$H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 \times \frac{3}{4}}{2g} = \frac{3v^2}{8g}$$

Now, let's find the ratio of the maximum heights:

$$\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}$$

The v² terms cancel out, and we get:

$$\frac{H_1}{H_2} = \frac{1}{3}$$

Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3