Question

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:

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

Hint:

To determine the ratio of maximum heights of projectiles, focus on the square of the sine of the angle of projection, \( \sin^2 \theta \), as height depends directly on it.

Answer 1

The Correct Option is C

The maximum height attained by a projectile can be derived from the vertical motion component of the projectile. The vertical velocity at the maximum height is zero, and the equation of motion is: \[ v_y^2 = u_y^2 - 2gH_{\text{max}}, \] where: \( u_y = u \sin \theta \) is the vertical component of the initial velocity, \( v_y = 0 \) at the maximum height, \( g \) is the acceleration due to gravity, \( H_{\text{max}} \) is the maximum height. Rearranging for \( H_{\text{max}} \), we get: \[ H_{\text{max}} = \frac{u_y^2}{2g} = \frac{(u \sin \theta)^2}{2g}. \] Let the two projectiles have angles of projection: \[ \theta_1 = 30^\circ \quad \text{and} \quad \theta_2 = 60^\circ. \] For the first projectile: \[ H_1 = \frac{(u \sin 30^\circ)^2}{2g}. \] For the second projectile: \[ H_2 = \frac{(u \sin 60^\circ)^2}{2g}. \] The ratio of their maximum heights is: \[ \frac{H_1}{H_2} = \frac{(u \sin 30^\circ)^2}{(u \sin 60^\circ)^2}. \] Canceling common terms \( u^2 \) and substituting \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we get: \[ \frac{H_1}{H_2} = \frac{\left(\frac{1}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2}. \] Simplify: \[ \frac{H_1}{H_2} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}. \] Thus, the ratio of the maximum heights is: \[ H_1 : H_2 = 1 : 3. \] Final Answer: \[ \boxed{1 : 3} \]