JEE Main 2025 Apr 7 Shift 1, Physics Q12: Motion In Two And Three Dimension

JEE Main2025Apr 7, Shift 1Physics

Two projectiles are fired from ground with same initial speeds from same point at angles $(4 5^{\circ} + α)$ and $(4 5^{\circ} - α)$ with horizontal direction. The ratio of their times of flights is

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

Solution

Time of flight for a projectile launched at angle $θ$ with speed $v$ : $T = \frac{2 v sin θ}{g}$ .

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

Expand using the sine addition formula:

$= \frac{\frac{1}{2} cos α + \frac{1}{2} sin α}{\frac{1}{2} cos α - \frac{1}{2} sin α} = \frac{cos α + sin α}{cos α - sin α}$

Divide numerator and denominator by $cos α$ :

$\frac{T _{1}}{T _{2}} = \frac{1 + tan α}{1 - tan α}$

$∴$ Option (4).

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JEE Main 2025 Apr 7 Shift 1, Physics Q12: Motion In Two And Three Dimension

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