Question

A body P is projected at an angle of 30° with the horizontal and another body Q is projected at angle of 30° with the vertical. If the ratio of the horizontal ranges of the bodies P and Q is 1:2, then the ratio of the maximum heights reached by the bodies P and Q is

• A. \(1:4\)
• B. \(1:6\)
• C. \(2:3\)
• D. \(1:1\)

Hint:

Use trigonometric identities and properties to analyze projectile motion, noting symmetry in angles.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the projectile motions of bodies P and Q and determine the ratio of their maximum heights based on the given information about their horizontal ranges. 1. Projectile Motion Basics: - The horizontal range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) with the horizontal is given by: \[ R = \frac{u^2 \sin(2\theta)}{g} \] - The maximum height \( H \) reached by the projectile is: \[ H = \frac{u^2 \sin^2(\theta)}{2g} \] 2. Given Information: - Body P is projected at an angle of 30° with the horizontal. - Body Q is projected at an angle of 30° with the vertical, which means it is projected at an angle of 60° with the horizontal. - The ratio of the horizontal ranges of P and Q is 1:2. 3. Calculate the Ranges: - For body P (\( \theta_P = 30° \)): \[ R_P = \frac{u_P^2 \sin(60°)}{g} = \frac{u_P^2 \cdot \frac{\sqrt{3}}{2}}{g} \] - For body Q (\( \theta_Q = 60° \)): \[ R_Q = \frac{u_Q^2 \sin(120°)}{g} = \frac{u_Q^2 \cdot \frac{\sqrt{3}}{2}}{g} \] - Given \( \frac{R_P}{R_Q} = \frac{1}{2} \): \[ \frac{\frac{u_P^2 \cdot \frac{\sqrt{3}}{2}}{g}}{\frac{u_Q^2 \cdot \frac{\sqrt{3}}{2}}{g}} = \frac{1}{2} \implies \frac{u_P^2}{u_Q^2} = \frac{1}{2} \] 4. Calculate the Maximum Heights: - For body P: \[ H_P = \frac{u_P^2 \sin^2(30°)}{2g} = \frac{u_P^2 \cdot \left(\frac{1}{2}\right)^2}{2g} = \frac{u_P^2}{8g} \] - For body Q: \[ H_Q = \frac{u_Q^2 \sin^2(60°)}{2g} = \frac{u_Q^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u_Q^2}{8g} \] - Using \( \frac{u_P^2}{u_Q^2} = \frac{1}{2} \): \[ H_P = \frac{u_P^2}{8g} = \frac{1}{2} \cdot \frac{u_Q^2}{8g} = \frac{u_Q^2}{16g} \] \[ H_Q = \frac{3u_Q^2}{8g} \] - Therefore, the ratio \( \frac{H_P}{H_Q} \) is: \[ \frac{H_P}{H_Q} = \frac{\frac{u_Q^2}{16g}}{\frac{3u_Q^2}{8g}} = \frac{1}{16} \cdot \frac{8}{3} = \frac{1}{6} \] 5. Final Answer: - The ratio of the maximum heights reached by bodies P and Q is: \[ \boxed{1:6} \] This corresponds to option (2).