Q 11-03-177JEE MainJEE Main 2025 (7 Apr, Shift 1)Easy
Two projectiles are fired from the ground with the same initial speeds from the same point at angles $(45^\circ + \alpha)$ and $(45^\circ - \alpha)$ with the horizontal. The ratio of their times of flight is:
Answer: (D) $\dfrac{1 + \tan\alpha}{1 - \tan\alpha}$
$T = \dfrac{2u\sin\theta}{g}$, so
$$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$
Solution by Sreeraj P, M.Sc Physics