Two projectiles are fired with same initial speed from same point on ground at angles of (45° – α) and (45° + α), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

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Two projectiles are fired with same initial speed from same point on ground at angles of (45° – $α$ ) and (45° + $α$ ), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

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$\frac{1 - \tan α}{1 + \tan α}$

$\frac{1 + \sin 2 α}{1 - \sin 2 α}$

$\frac{1 + sinα}{1 - sinα}$

$\frac{1 - \sin 2 α}{1 + \sin 2 α}$

answer is C.

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Detailed Solution

$\begin{array}{l} H_{Max} = \frac{(usin θ)^{2}}{2 g} \\ \frac{{(H_{max})}_{1}}{{(H_{max})}_{2}} = \frac{u^{2} \sin^{2} (45 - α)}{u^{2} \sin^{2} (45 + α)} \\ = \frac{{(\frac{1}{\sqrt{2}} \cos α - \frac{1}{\sqrt{2}} \sin α)}^{2}}{{(\frac{1}{\sqrt{2}} \cos α + \frac{1}{\sqrt{2}} \sin α)}^{2}} \\ = \frac{1 - \sin 2 α}{1 + \sin 2 α} \end{array}$