Question:
A projectile is fired from the surface of the earth with a velocity of \(v_0\) at an angle \(\theta\) with the horizontal. If the maximum height reached is equal to its horizontal range, then the angle of projection \(\theta\) is given by:
A projectile is fired from the surface of the earth with a velocity of \(v_0\) at an angle \(\theta\) with the horizontal. If the maximum height reached is equal to its horizontal range, then the angle of projection \(\theta\) is given by:
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Memorize the relation \(\tan\theta = \frac{4H}{R}\) for quick objective problems. If \(H = R\), then \(\tan\theta\) is always equal to \(4\).
Updated On: May 25, 2026
- \(\tan^{-1}(1)\)
- \(\tan^{-1}(2)\)
- \(\tan^{-1}(4)\)
- \(\tan^{-1}(0.5)\)
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The Correct Option is C
Solution and Explanation
Concept:
For any projectile launched with an initial velocity \(v_0\) at an angle \(\theta\) with the horizontal under gravity \(g\), the formulas for maximum height (\(H\)) and horizontal range (\(R\)) are given by:
\[
H = \frac{v_0^2 \sin^2\theta}{2g}
\]
\[
R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{2v_0^2 \sin\theta \cos\theta}{g}
\]
We can establish a direct geometric relationship between the height and range by finding their ratio.
Step 1: Deriving the fundamental relation between \(H\) and \(R\).
Dividing the maximum height equation by the horizontal range equation: \[ \frac{H}{R} = \frac{\frac{v_0^2 \sin^2\theta}{2g}}{\frac{2v_0^2 \sin\theta \cos\theta}{g}} \] Simplifying the fractions by cancelling out the common terms \(v_0^2\) and \(g\): \[ \frac{H}{R} = \frac{\sin^2\theta}{2 \cdot 2 \sin\theta \cos\theta} = \frac{\sin\theta}{4 \cos\theta} = \frac{\tan\theta}{4} \] This gives us the standard projectile identity: \[ \tan\theta = \frac{4H}{R} \]
Step 2: Applying the problem's condition.
The problem states that the maximum height reached is equal to the horizontal range, meaning \(H = R\). Substituting this condition into our identity: \[ \tan\theta = \frac{4H}{H} = 4 \] Taking the inverse tangent on both sides to solve for the angle of projection \(\theta\): \[ \theta = \tan^{-1}(4) \]
Dividing the maximum height equation by the horizontal range equation: \[ \frac{H}{R} = \frac{\frac{v_0^2 \sin^2\theta}{2g}}{\frac{2v_0^2 \sin\theta \cos\theta}{g}} \] Simplifying the fractions by cancelling out the common terms \(v_0^2\) and \(g\): \[ \frac{H}{R} = \frac{\sin^2\theta}{2 \cdot 2 \sin\theta \cos\theta} = \frac{\sin\theta}{4 \cos\theta} = \frac{\tan\theta}{4} \] This gives us the standard projectile identity: \[ \tan\theta = \frac{4H}{R} \]
Step 2: Applying the problem's condition.
The problem states that the maximum height reached is equal to the horizontal range, meaning \(H = R\). Substituting this condition into our identity: \[ \tan\theta = \frac{4H}{H} = 4 \] Taking the inverse tangent on both sides to solve for the angle of projection \(\theta\): \[ \theta = \tan^{-1}(4) \]
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