A stone projected with a velocity u at an angle $θ$ with the horizontal reaches maximum height $H_{1}$ . When it is projected with velocity u at an angle $(\frac{π}{2} - θ)$ with the horizontal reaches maximum height $H_{2}$ . The relation between horizontal range, R of the projectile, $H_{1}$ and $H_{2}$ is

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

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

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

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

answer is A.

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Detailed Solution

Height reached by the stone 1^st time

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

Height reached by the stone 2^nd time

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

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

But range $R = \frac{u^{2} (2 \sin θ \cos θ)}{g}$

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

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