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

Correct answer: B. ((1- 2 α)/(1+ 2 α)) aligned & H_ Max = (u θ)^2 2 ~g & (H_ )_1 (H_ )_2 = u^2 ^2(45-α) u^2 ^2(45+α) & = ( 1 2 α- 1 2 α)^2 ( 1 2 α+ 1 2 α)^2 & =(1- 2 α)/(1+ 2 α) aligned

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2025 · 29 January · Morning Shift · Physics · Motion in Two Dimensions

Two projectiles are fired with same initial speed from same point...

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 :

Options

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})

Correct answer

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

Step-by-step solution

\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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