Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

<p>Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:</p> <p>$H = \frac{v^2 \sin^2 \theta}{2g}$</p> <p>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.</p> <p>For the projectile projected at 30°, the maximum height is:</p> <p>$$H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 \times \frac{1}{4}}{2g} = \frac{v^2}{8g}$$</p> <p>For the projectile projected at 60°, the maximum height is:</p> <p>$$H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 \times \frac{3}{4}}{2g} = \frac{3v^2}{8g}$$</p> <p>Now, let&#39;s find the ratio of the maximum heights:</p> <p>$\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}$</p> <p>The v² terms cancel out, and we get:</p> <p>$\frac{H_1}{H_2} = \frac{1}{3}$</p> <p>Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3</p>