Question:medium

Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is $\pi/8$ and the maximum height reached by it is $102$ metres. Then the maximum height reached by the other in metres is

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For complementary angles $\theta$ and $(90-\theta)$, the ratio of maximum heights is $H_1/H_2 = \tan^2 \theta$.
  • $336$
  • $224$
  • $56$
  • $34$
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The Correct Option is D

Solution and Explanation

Step 1: Identify the Angles: First angle $\theta_1 = \pi/8 = 22.5^\circ$. Second angle $\theta_2 = \pi/2 - \pi/8 = 3\pi/8 = 67.5^\circ$.

Step 2: Relate Heights to Angles: The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$. Since $u$ and $g$ are constant, $H \propto \sin^2 \theta$. $$\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2} = \frac{\sin^2(\pi/8)}{\sin^2(3\pi/8)}$$ Note that $\sin(3\pi/8) = \cos(\pi/8)$, so: $$\frac{H_1}{H_2} = \frac{\sin^2(\pi/8)}{\cos^2(\pi/8)} = \tan^2(\pi/8)$$

Step 3: Calculate $\tan^2(\pi/8)$: Using the identity $\tan^2(\theta/2) = \frac{1 - \cos \theta}{1 + \cos \theta}$ for $\theta = \pi/4$: $$\tan^2(\pi/8) = \frac{1 - \cos(\pi/4)}{1 + \cos(\pi/4)} = \frac{1 - 1/\sqrt{2}}{1 + 1/\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$$ Multiplying by conjugate: $(\sqrt{2}-1)^2 = 3 - 2\sqrt{2} \approx 0.1715$.
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