Kinematics

JEE Main level, single correct, medium

209 of 308

Question 209 of 308

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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Answer: (B) \(\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}\)

HMax=(usin⁡θ)22 g(Hmax⁡)1(Hmax⁡)2=u2sin⁡2(45−α)u2sin⁡2(45+α)=(12cos⁡α−12sin⁡α)2(12cos⁡α+12sin⁡α)2=1−sin⁡2α1+sin⁡2α\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)} \\ & =\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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