A particle is projected from ground at some angle with the horizontal. Let P be the point at maximum height H. At what height above the point P should the particle be aimed to have range equal to maximum height?

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H/2

answer is A.

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Detailed Solution

We find that $H = R or \frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2} 2 \sin θ \cos θ}{g}$

$⇒ \tan θ = 4 or θ = \tan^{- 1} (4)$

$A C = R / 2, P C = H$

$\tan θ = 4$

Now $\tan θ = \frac{M C}{A C} = \frac{M P + P C}{A C}$

$\begin{matrix} 4 = \frac{h + H}{R / 2} \Rightarrow 4 = \frac{(h + H) 2}{H} \\ \Rightarrow h = H \end{matrix}$

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