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 (a) (1+tan α)/(1-tan α) (b) (1+sin 2α)/(1-sin 2α) (c) (1-tan α)/(1+tan α) (d) 1

Answer: (a). T = (2usin θ)/(g), so (T₁)/(T₂) = (sin (45° + α))/(sin (45° - α)) = (cos α + sin α)/(cos α - sin α) (the 1/√(2) cancels) Divide by cos α: (1 + tan α)/(1 - tan α) So the answer is (a).

Two projectiles are fired from ground with same initial speeds from… | JEE Main 2025 (7 Apr, Shift 1)

Q8JEE Main 2025 · 7 APR · SHIFT 1Easy

Comparing Two Projectiles

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

Show answer and solution

Answer: (a)

$T = \frac{2u\sin\theta}{g}$ , so $\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$
$= \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$ (the $\frac{1}{\sqrt{2}}$ cancels)
Divide by $\cos\alpha$ : $\frac{1 + \tan\alpha}{1 - \tan\alpha}$
So the answer is (a).

Report a mistake

Comparing Two Projectiles question from JEE Main 2025 (7 Apr, Shift 1)

More questions on Comparing Two Projectiles