Question

An object is projected such that it has to attain maximum range, while another body is projected to reach maximum height. If both objects reached the same maximum height, then find the ratio of their initial velocities.

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

Hint:

For maximum range, project at \( 45^\circ \). For maximum height, project vertically. Use the height formula \( H = \frac{u^2 \sin^2 \theta}{2g} \) to compare cases.

Answer 1

The Correct Option is B

Step 1: Understanding Maximum Height Condition For a projectile, the maximum height attained is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. 

Step 2: Maximum Range Projection For maximum range, the projectile is launched at \( 45^\circ \), so the height attained is: \[ H_R = \frac{u_R^2 \sin^2 45^\circ}{2g} = \frac{u_R^2}{4g} \] 

Step 3: Maximum Height Projection For maximum height, the projectile is launched vertically (\(\theta = 90^\circ\)), so the height attained is: \[ H_H = \frac{u_H^2}{2g} \] Since both objects attain the same height, \[ \frac{u_R^2}{4g} = \frac{u_H^2}{2g} \] Solving for \( \frac{u_R}{u_H} \): \[ \frac{u_R^2}{u_H^2} = 2 \quad \Rightarrow \quad \frac{u_R}{u_H} = \sqrt{2}:1 \]