A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H₁ . When it is projected with velocity u at an angle $(\frac{π}{2} - θ)$ with the horizontal, it reaches maximum height H₂ . The relation between the horizontal range R of the projectile, heights H₁ and H₂ is

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$R = 4 \sqrt{H_{1} H_{2}}$

$R = 4 (H_{1} + H_{2})$

$R = 4 (H_{1} - H_{2})$

$R = \frac{{H_{1}}^{2}}{{H_{2}}^{2}}$

answer is A.

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Detailed Solution

$H_{1} = \frac{u^{2} \sin^{2} θ}{2 g}$

and $H_{2} = \frac{u^{2} \sin^{2} (90^{ο} - θ)}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g}$

$H_{1} H_{2} = \frac{u^{2} \sin^{2} θ}{2 g} \times \frac{u^{2} \cos^{2} θ}{2 g} = \frac{{(u^{2} \sin 2 θ)}^{2}}{16 g^{2}} = \frac{R^{2}}{16}$

$∴ R = 4 \sqrt{H_{1} H_{2}}$

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