If H₁ and H₂ be the greatest heights of a projectile in two paths for a given value of range, then the horizontal range of projectile is given by

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

$\frac{H_{1} + H_{2}}{2}$

$\frac{H_{1} + H_{2}}{4}$

$4 [H_{1} + H_{2}]$

answer is C.

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Detailed Solution

$θ_{1} + θ_{2} = 90^{\circ}$

$\begin{array}{l} H_{1} = \frac{u^{2} \sin^{2} θ_{1}}{2 g} & H_{2} = \frac{u^{2} \sin (90^{\circ} - θ_{1})}{2 g} \\ H_{1} H_{2} = \frac{R^{2}}{16} ∵ R = \frac{u^{2} \sin 2 θ_{1}}{g} \Rightarrow R = 4 \sqrt{H_{1} H_{2}} \end{array}$

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