Two projectiles are fired from ground with same initial speeds from same point at angles (45^ +α) and (45^ -α) with horizontal direction. The ratio of their times of flights is

Correct answer: D. (1+ α)/(1- α) θ_1=45+α ; θ_2=45-α Time of flight, T= 2 v θ g T_1 ~T_2 =( (45+α))/( (45-α)) T_1 ~T_2 = 1 2 α+ 1 2 α 1 2 α- 1 2 α T_1 ~T_2 =( α+ α)/( α- α)=(1+ α)/(1- α)

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2025 · 7 April · Morning Shift · Physics · Motion in Two Dimensions

Two projectiles are fired from ground with same initial speeds from...

Two projectiles are fired from ground with same initial speeds from same point at angles

\left(45^{\circ}+\alpha\right)

and

\left(45^{\circ}-\alpha\right)

with horizontal direction. The ratio of their times of flights is

Options

A1
B $\frac{1-\tan \alpha}{1+\tan \alpha}$
C $\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}$
D $\frac{1+\tan \alpha}{1-\tan \alpha}$

Correct answer

D. $\frac{1+\tan \alpha}{1-\tan \alpha}$

Step-by-step solution

\theta_1=45+\alpha ; \theta_2=45-\alpha

Time of flight,

\mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}

\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}

\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha}{\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha}

\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}

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