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A projectile aimed at a mark which is in the horizontal plane through the point of projection falls “a” distance short of it when the elevation is $α$ and goes “b” distance too far when the elevation is $β$ . If the velocity of projection is same in all the cases, the proper elevation is

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$\frac{1}{2} \sin^{- 1} [\frac{a \sin 2 α - b \sin 2 β}{a - b}]$

$\frac{1}{2} \sin^{- 1} [\frac{asin 2 α + bsin 2 β}{a - b}]$

$\frac{1}{2} \sin^{- 1} [\frac{a \sin 2 α + b \sin 2 β}{a + b}]$

$\frac{1}{2} \sin^{- 1} [\frac{a \sin 2 α - b \sin 2 β}{a + b}]$

answer is B.

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Detailed Solution

Let ‘ $θ$ ’ be the proper elevation $a + \frac{u^{2}}{g} \sin 2 α = \frac{u^{2}}{g} . \sin 2 θ$

$\frac{u^{2}}{g} \sin 2 f - b = \frac{u^{2}}{g} \sin 2 θ$

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