JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 8)
Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :
$ \frac{1+\sin\alpha}{1-\sin\alpha} $
$ \frac{1+\sin2\alpha}{1-\sin2\alpha} $
$ \frac{1-\tan\alpha}{1+\tan\alpha} $
$ \frac{1-\sin2\alpha}{1+\sin2\alpha} $
Explanation
$$\begin{aligned}
& \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{usin} \theta)^2}{2 \mathrm{~g}} \\
& \frac{\left(\mathrm{H}_{\max }\right)_1}{\left(\mathrm{H}_{\max }\right)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \\
& =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^2} \\
& =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}
\end{aligned}$$
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