Question:
The percentage decrease in range of a projectile projected at $30^\circ$ when compared to maximum range is:
The percentage decrease in range of a projectile projected at $30^\circ$ when compared to maximum range is:
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$\sin(60^\circ)$ is $\sqrt{3}/2 \approx 0.866$; the range is $86.6\%$ of the maximum, meaning a $13.4\%$ decrease.
Updated On: Jun 10, 2026
- 13.4
- 25
- 50
- 75
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Projectile range $R = \frac{u^2 \sin(2\theta)}{g}$.
Step 2: Analysis
Maximum range $R_{max} = \frac{u^2}{g}$ (at $\theta = 45^\circ$). Range at $30^\circ$: $R_{30} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \sqrt{3}/2}{g} \approx 0.866 R_{max}$. Percentage decrease $= \frac{R_{max} - 0.866 R_{max}}{R_{max}} \times 100 \approx 13.4\%$.
Step 3: Conclusion
The decrease is approximately 13.4
Final Answer: (A)
Projectile range $R = \frac{u^2 \sin(2\theta)}{g}$.
Step 2: Analysis
Maximum range $R_{max} = \frac{u^2}{g}$ (at $\theta = 45^\circ$). Range at $30^\circ$: $R_{30} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \sqrt{3}/2}{g} \approx 0.866 R_{max}$. Percentage decrease $= \frac{R_{max} - 0.866 R_{max}}{R_{max}} \times 100 \approx 13.4\%$.
Step 3: Conclusion
The decrease is approximately 13.4
Final Answer: (A)
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