JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Kinematics

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✓ Correct

Answer: (D)

Solution

For a projectile,

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

, hence

tan θ = \frac{4}{3}

, so

sin θ = \frac{4}{5}

and

cos θ = \frac{3}{5}

. Substituting into the range,

R = \frac{u ^{2} 2 \cdot \frac{4}{5} \cdot \frac{3}{5}}{g} = \frac{24 u ^{2}}{25 g}

, so the value of

n

24

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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