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

(D) $\frac{1+\tan \alpha}{1-\tan \alpha}$

(B) $\frac{1-\tan \alpha}{1+\tan \alpha}$

(C) $\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}$

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Kinematics

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

4 5^{\circ} + α

) and (

4 5^{\circ} - α

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

(A)1
(B) $\frac{1 - t a n α}{1 + t a n α}$
(C) $\frac{1 + s i n 2 α}{1 - s i n 2 α}$
(D) $\frac{1 + t a n α}{1 - t a n α}$ ✓ Correct

Answer: (D)

Solution

The time of flight is

T = \frac{2 u sin θ}{g}

, so for the two launch angles the ratio depends only on the sines:

\frac{T _{1}}{T _{2}} = \frac{sin ( 4 5 ^{\circ} + α )}{sin ( 4 5 ^{\circ} - α )} = \frac{cos α + sin α}{cos α - sin α} .

Dividing numerator and denominator by

cos α

\frac{T _{1}}{T _{2}} = \frac{1 + tan α}{1 - tan α} .

The ratio of their times of flight is

\frac{1 + tan α}{1 - tan α}

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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