Question:
Two objects A and B are projected with same velocity at angles \( \theta \) and \( 90-\theta \) respectively with the horizontal. Then the ratio of maximum heights they reached \( \frac{H_A}{H_B} \) is:
Two objects A and B are projected with same velocity at angles \( \theta \) and \( 90-\theta \) respectively with the horizontal. Then the ratio of maximum heights they reached \( \frac{H_A}{H_B} \) is:
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Projectile motion with complementary angles results in the same range but different maximum heights related by the square of the tangent of the angle.
Updated On: Jun 9, 2026
- \( \tan \theta \)
- \( \tan^2 \theta \)
- \( 2 \tan \theta \)
- \( \cot^2 \theta \)
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The Correct Option is B
Solution and Explanation
Concept:
The vertical component of a projectile's velocity is \( v_y = v \sin \theta \). The maximum height \( H \) is derived from the conservation of energy or kinematic equations: \( H = \frac{v_y^2}{2g} = \frac{v^2 \sin^2 \theta}{2g} \).
Step 1: Express maximum heights for both projectiles.
For object A projected at angle \(\theta\): $$ H_A = \frac{v^2 \sin^2 \theta}{2g} $$ For object B projected at angle \(90^\circ - \theta\): $$ H_B = \frac{v^2 \sin^2(90^\circ - \theta)}{2g} $$
Step 2: Apply trigonometric identity.
Using the identity \(\sin(90^\circ - \theta) = \cos \theta\): $$ H_B = \frac{v^2 \cos^2 \theta}{2g} $$
Step 3: Form the ratio.
$$ \frac{H_A}{H_B} = \frac{\frac{v^2 \sin^2 \theta}{2g}}{\frac{v^2 \cos^2 \theta}{2g}} $$ $$ \frac{H_A}{H_B} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta $$ $$\boxed{\tan^2 \theta}$$
Step 1: Express maximum heights for both projectiles.
For object A projected at angle \(\theta\): $$ H_A = \frac{v^2 \sin^2 \theta}{2g} $$ For object B projected at angle \(90^\circ - \theta\): $$ H_B = \frac{v^2 \sin^2(90^\circ - \theta)}{2g} $$
Step 2: Apply trigonometric identity.
Using the identity \(\sin(90^\circ - \theta) = \cos \theta\): $$ H_B = \frac{v^2 \cos^2 \theta}{2g} $$
Step 3: Form the ratio.
$$ \frac{H_A}{H_B} = \frac{\frac{v^2 \sin^2 \theta}{2g}}{\frac{v^2 \cos^2 \theta}{2g}} $$ $$ \frac{H_A}{H_B} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta $$ $$\boxed{\tan^2 \theta}$$
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