Question:
A projectile is launched from the ground with an initial velocity $v$ at an angle $\theta$ to the horizontal. If its horizontal range is equal to its maximum height, then the value of $\tan \theta$ is:
A projectile is launched from the ground with an initial velocity $v$ at an angle $\theta$ to the horizontal. If its horizontal range is equal to its maximum height, then the value of $\tan \theta$ is:
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Use the direct relationship between height and range: $\tan\theta = \frac{4H}{R}$. Setting $H = R$ instantly yields $\tan\theta = 4$.
Updated On: Jun 3, 2026
- $4$
- $2$
- $1$
- $\frac{1}{4}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The maximum height $H$ reached by a projectile is given by $H = \frac{v^2 \sin^2 \theta}{2g}$, and the horizontal range $R$ is given by $R = \frac{v^2 \sin 2\theta}{g}$.
Step 2: Meaning
We are given that the horizontal range is equal to the maximum height ($R = H$). We equate their formulas to solve for $\theta$.
Step 3: Analysis
Equating $H$ and $R$: \[ \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin 2\theta}{g} \] Using the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ and simplifying: \[ \frac{\sin^2 \theta}{2} = 2\sin\theta\cos\theta \] Assuming $\sin\theta \neq 0$ (as the projectile is launched at a non-zero angle): \[ \frac{\sin\theta}{2} = 2\cos\theta \implies \frac{\sin\theta}{\cos\theta} = 4 \implies \tan\theta = 4 \]
Step 4: Conclusion
The value of $\tan\theta$ is $4$.
Final Answer: (A)
The maximum height $H$ reached by a projectile is given by $H = \frac{v^2 \sin^2 \theta}{2g}$, and the horizontal range $R$ is given by $R = \frac{v^2 \sin 2\theta}{g}$.
Step 2: Meaning
We are given that the horizontal range is equal to the maximum height ($R = H$). We equate their formulas to solve for $\theta$.
Step 3: Analysis
Equating $H$ and $R$: \[ \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin 2\theta}{g} \] Using the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ and simplifying: \[ \frac{\sin^2 \theta}{2} = 2\sin\theta\cos\theta \] Assuming $\sin\theta \neq 0$ (as the projectile is launched at a non-zero angle): \[ \frac{\sin\theta}{2} = 2\cos\theta \implies \frac{\sin\theta}{\cos\theta} = 4 \implies \tan\theta = 4 \]
Step 4: Conclusion
The value of $\tan\theta$ is $4$.
Final Answer: (A)
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