Question:medium

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:

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For complementary angles, \(\sin \theta\) and \(\cos \theta\) swap — use identities quickly!
Updated On: Apr 17, 2026
  • \(1:1 \)
  • \(1 : \tan \theta \)
  • \(\tan \theta : 1 \)
  • \(\tan^2 \theta : 1 \)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The maximum height of a projectile depends on its vertical component of initial velocity.
When two projectiles are thrown with the same speed but at different angles, their heights will differ based on the sine of their respective projection angle(s).
Step 2: Key Formula or Approach:
The maximum height \(H\) attained by a projectile is given by the formula:
\[ H = \frac{u^{2} \sin^{2} \alpha}{2g} \]
Where \(u\) is the initial velocity, \(\alpha\) is the angle of projection, and \(g\) is the acceleration due to gravity.
Step 3: Detailed Explanation:
Let the first stone be projected at an angle \(\alpha_{1} = \theta\).
Its maximum height \(H_{1}\) is:
\[ H_{1} = \frac{u^{2} \sin^{2} \theta}{2g} \dots (i) \]
The second stone is projected at an angle \(\alpha_{2} = 90^{\circ} - \theta\).
Its maximum height \(H_{2}\) is:
\[ H_{2} = \frac{u^{2} \sin^{2} (90^{\circ} - \theta)}{2g} \]
Using the trigonometric identity \(\sin (90^{\circ} - \theta) = \cos \theta\), we get:
\[ H_{2} = \frac{u^{2} \cos^{2} \theta}{2g} \dots (ii) \]
Now, find the ratio of \(H_{1}\) to \(H_{2}\):
\[ \frac{H_{1}}{H_{2}} = \frac{\frac{u^{2} \sin^{2} \theta}{2g}}{\frac{u^{2} \cos^{2} \theta}{2g}} = \frac{\sin^{2} \theta}{\cos^{2} \theta} \]
\[ \frac{H_{1}}{H_{2}} = \tan^{2} \theta \]
Thus, the ratio is \(\tan^{2} \theta : 1\).
Step 4: Final Answer:
The ratio of the maximum heights is \(\tan^{2} \theta : 1\).
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