Two projectiles are fired with same initial speed from same point… - Vidyadip

MCQ

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 :

A
$\frac{1-\tan \alpha}{1+\tan \alpha}$
B
$\frac{1+\sin \alpha}{1-\sin \alpha}$
✓
$\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$
D
$\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}$

✓

Answer

Correct option: C.

$\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$

(C)
Sol.
$H_{M a x}=\frac{(u \sin \theta)^{2}}{2 g}$
$\frac{\left(\mathrm{H}_{\max }\right)_{1}}{\left(\mathrm{H}_{\max }\right)_{2}}=\frac{\mathrm{u}^{2} \sin ^{2}(45-\alpha)}{\mathrm{u}^{2} \sin ^{2}(45+\alpha)}$
$=\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^{2}}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^{2}}$
$=\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$

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