JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}\)
- C \(\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}\)
- D \(\frac{1+\sin \alpha}{1-\sin \alpha}\)
Answer & Solution
Correct Answer
(B) \(\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{u} \sin \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)} \\ &…
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