Two projectiles are fired with same initial speed from same point on ground at angles of (45^°-α ) and (45^°+α ) , respectively, with the horizontal direction. The ratio of their maximum heights attained is :

Correct answer: (B) (1-sin 2 α)/(1+sin 2 α). H_Max=((u sin θ)^2)/(2 ~g) frac (H_max )_1 (H_max )_2=(u^2 sin^2(45-α))/(u^2 sin^2(45+α)) =((1/(√2) cos α-1/(√2) sin α )^2)/((1/(√2) cos α+1/(√2) sin α )^2) =(1-sin 2 α)/(1+sin 2 α)

Q56 / 75 · JEE Main 2025 (29 Jan Shift 1) · Physics

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

A
$\frac{1-\tan \alpha}{1+\tan \alpha}$
Correct answer
B
$\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$
C
$\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}$
D
$\frac{1+\sin \alpha}{1-\sin \alpha}$

Solution

$\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{u} \sin \theta)^2}{2 \mathrm{~g}} \\ & \frac{\left(\mathrm{H}_{\max }\right)_1}{\left(\mathrm{H}_{\max }\right)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \\ & =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^2} \\ & =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}\end{aligned}$

Hence, the correct answer is option: 'B'.

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

Solution

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