Two projectiles are fired with same initial speed from same point on ground at angles of $\left(45^{\circ}-\alpha\right)$ and ( $45^{\circ}+\alpha$ ), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

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

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

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

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

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Kinematics

Two projectiles are fired with same initial speed from same point on ground at angles of

(4 5^{\circ} - α)

and (

4 5^{\circ} + α

), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

Answer: (C)

Maximum height of a projectile launched at angle

θ

with speed

u

H = \frac{u ^{2} sin ^{2} θ}{2 g}

For the two projectiles, with angles

(45° - α)

and

(45° + α)

and the same

u

\frac{H _{1}}{H _{2}} = \frac{sin ^{2} ( 45° - α )}{sin ^{2} ( 45° + α )}

Using

sin (45° \mp α) = \frac{1}{2} (cos α \mp sin α)

\frac{H _{1}}{H _{2}} = \frac{( cos α - sin α ) ^{2}}{( cos α + sin α ) ^{2}} = \frac{1 - 2 sin α cos α}{1 + 2 sin α cos α} = \frac{1 - sin 2 α}{1 + sin 2 α}

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option