JEE Main 2025 Apr 3 Shift 2, Physics Q11: Motion In Two And Three Dimension

JEE Main2025Apr 3, Shift 2Physics

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u ^{2}}{25 g}$ , where value of $n$ is :

(Given ' $g$ ' is the acceleration due to gravity).

A
$6$
B
$18$
C
$12$
D
$24$

Solution

For projection angle $θ$ , $R = \frac{u ^{2} sin 2 θ}{g}$ and $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . The condition $R = 3 H$ gives

$2 sin θ cos θ = \frac{3}{2} sin^{2} θ \Rightarrow tan θ = \frac{4}{3}$ .

So $sin θ = 4/5$ , $cos θ = 3/5$ , and $sin 2 θ = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$ .

$R = \frac{u ^{2}}{g} \cdot \frac{24}{25} = \frac{24 u ^{2}}{25 g}$ , hence $n = 24$ .

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JEE Main 2025 Apr 3 Shift 2, Physics Q11: Motion In Two And Three Dimension

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