Q 11-03-056JEE MainJEE Main 2025 (29 Jan, Shift 1)Easy
Two projectiles are fired with the same initial speed from the same point on the ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is
Answer: (B) $\dfrac{1 - \sin2\alpha}{1 + \sin2\alpha}$
$H = \dfrac{u^2\sin^2\theta}{2g} \propto \sin^2\theta = \dfrac{1 - \cos2\theta}{2}$.
$$\frac{H_1}{H_2} = \frac{1 - \cos(90^\circ - 2\alpha)}{1 - \cos(90^\circ + 2\alpha)} = \frac{1 - \sin2\alpha}{1 + \sin2\alpha}$$
Solution by Sreeraj P, M.Sc Physics