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Two balls A and B are thrown with speeds u and $\frac{u}{2}$ respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of ball B is $15^{0}$ with the horizontal, then the angle of projection of A is:

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$\frac{1}{2} \sin^{- 1} (\frac{1}{8})$

$\frac{1}{4} \sin^{- 1} (\frac{1}{8})$

$\frac{1}{3} \sin^{- 1} (\frac{1}{8})$

$\sin^{- 1} (\frac{1}{8})$

answer is B.

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Detailed Solution

$\frac{u^{2} \sin 2 θ}{g} = \frac{{(\frac{u}{2})}^{2} \sin 30^{0}}{g} = \frac{u^{2}}{8 g}$

$\sin 2 θ = \frac{1}{8}$

or $θ = \frac{1}{2} \sin^{- 1} (\frac{1}{8})$

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