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Galileo writes that for angles of projection of a projectile at angles (45^o + $θ$ ) and (45^o - $θ$ ), the horizontal ranges described by the projectile are in the ratio of

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2 : 1

2 : 3

1 : 2

1 : 1

answer is C.

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Detailed Solution

$R_{1} = \frac{u^{2} \sin (90^{\circ} - 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$

$R_{2} = \frac{u^{2} \sin (90^{\circ} + 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$

$\Rightarrow R_{1} = R_{2}$

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