A projectile is aimed at a mark on the horizontal plane through the point of projection and falls 12m short when the angle of projection is 15°, while it overshoots the mark by 24m when the angle of projection is 45°.The angle of projection to hit the mark is

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$\sin^{- 1} (\frac{2}{3})$

$\frac{1}{2} \sin^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{2}{3})$

2 $\sin^{- 1} (\frac{2}{3})$

answer is C.

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Detailed Solution

$\begin{array}{l} R - 12 = \frac{u^{2} \sin 30}{g} = \frac{u^{2}}{2 g} \\ R + 24 = \frac{u^{2} \sin 90}{g} = \frac{u^{2}}{g} \Rightarrow \frac{R - 12}{R + 24} = \frac{1}{2} \end{array}$
Solve and get the value of R substitute R in one of the expression and find ' $θ$ '

R = 48,

$\Rightarrow \frac{u^{2}}{g} = R + 24 = 72$
$∵ R = \frac{u^{2} \sin 2 θ}{g} \Rightarrow \sin 2 θ = 2 / 3$

$θ = \frac{1}{2} \sin^{- 1} (2 / 3)$

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