JEE Main 2025 Jan 29 Shift 1, Physics Q2: Motion In Two And Three Dimension

JEE Main2025Jan 29, Shift 1Physics

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 :

A
$\frac{1 - tan α}{1 + tan α}$
B
$\frac{1 + sin α}{1 - sin α}$
C
$\frac{1 - sin 2 α}{1 + sin 2 α}$
D
$\frac{1 + sin 2 α}{1 - sin 2 α}$

Solution

For a projectile with launch speed $u$ at angle $θ$ , $H_{m a x} = \frac{u ^{2} sin ^{2} θ}{2 g} .$ The ratio for the two angles is

$\frac{H _{1}}{H _{2}} = \frac{sin ^{2} ( 4 5 ^{\circ} - α )}{sin ^{2} ( 4 5 ^{\circ} + α )} .$

Use $sin (4 5^{\circ} \pm α) = \frac{1}{2} (cos α \pm sin α)$ and square:

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

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JEE Main 2025 Jan 29 Shift 1, Physics Q2: Motion In Two And Three Dimension

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