Question

Two projectiles are projected at 30° and 60° with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is :

• A. \(1:\sqrt 3\)
• B. 1:3
• C. \(2:\sqrt 3\)
• D. \(\sqrt 3:1\)

Answer 1

The Correct Option is B

In projectile motion, the maximum height \( H \) attained by a projectile is given by the formula: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. At \( \theta = 30^\circ \), the maximum height \( H_1 \) is: \[ H_1 = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g} \] At \( \theta = 60^\circ \), the maximum height \( H_2 \) is: \[ H_2 = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u^2}{8g} \] Now, taking the ratio of the maximum heights: \[ \frac{H_1}{H_2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3} \] Thus, the ratio is \( 1 : 3 \).