Question

Two stones having different masses \( m_1 \) and \( m_2 \) are projected at angles \( \theta \) and \( (90^\circ - \theta) \) with same velocity from the same point. The ratio of their maximum heights is:

• A. \(1:1 \)
• B. \(1 : \tan \theta \)
• C. \(\tan \theta : 1 \)
• D. \(\tan^2 \theta : 1 \)

Hint:

For complementary angles, \(\sin \theta\) and \(\cos \theta\) swap — use identities quickly!

Answer 1

The Correct Option is D

Concept: Maximum height of projectile: \[ H = \frac{u^2 \sin^2 \theta}{2g} \]

Step 1: Heights of two projectiles: \[ H_1 = \frac{u^2 \sin^2 \theta}{2g}, \quad H_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} \]

Step 2: Using identity: \[ \sin (90^\circ - \theta) = \cos \theta \] \[ H_2 = \frac{u^2 \cos^2 \theta}{2g} \]

Step 3: Ratio: \[ \frac{H_1}{H_2} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \] \[ \therefore H_1 : H_2 = \tan^2 \theta : 1 \]