A stone projected with a velocity u angle θ with the horizontal reaches maximum height H1. When it is projected with velocity u at an angle π2−θ with the horizontal, it reaches maximum height H2. The relation between the horizontal range R of the projectile, H1 and H2 is

H1=u2sin2⁡θ2g and H2=u2sin2⁡(90−θ)2g=u2cos2⁡θ2gH1H2=u2sin2⁡θ2g×u2cos2⁡θ2g=u2sin⁡2θ216g2=R216∴ R=4H1H2

A stone projected with a velocity u angle θ with the horizontal reaches maximum height H1. When it is projected with velocity u at an angle π2−θ with the horizontal, it reaches maximum height H2. The relation between the horizontal range R of the projectile, H1 and H2 is

H1=u2sin2⁡θ2g and H2=u2sin2⁡(90−θ)2g=u2cos2⁡θ2gH1H2=u2sin2⁡θ2g×u2cos2⁡θ2g=u2sin⁡2θ216g2=R216∴ R=4H1H2

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A stone projected with a velocity u 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, 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

$\begin{array}{l} H_{1} = \frac{u^{2} \sin^{2} θ}{2 g} and H_{2} = \frac{u^{2} \sin^{2} (90 - θ)}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g} \\ H_{1} H_{2} = \frac{u^{2} \sin^{2} θ}{2 g} \times \frac{u^{2} \cos^{2} θ}{2 g} = \frac{{(u^{2} \sin 2 θ)}^{2}}{16 g^{2}} = \frac{R^{2}}{16} \\ ∴ R = 4 \sqrt{H_{1} H_{2}} \end{array}$