Two projectiles are fired with same initial speed from same point on ground at angles of $\left(45^{\circ}-\alpha\right)$ and $\left(45^{\circ}+\alpha\right)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{u} \sin 	heta)^2}{2 \mathrm{~g}} \ & \frac{\left(\mathrm{H}_{\max }ight)_1}{\left(\mathrm{H}_{\max }ight)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \ & =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alphaight)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alphaight)^2} \ & =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}\end{aligned}$

Two projectiles are fired with same initial speed from same point on ground at angles of…

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