JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Work, Power & Energy

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A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

(A) $\frac{S}{2}, \frac{3 gS}{2}$
(B) $\frac{S}{2}, \frac{3 gS}{2}$
(C) $\frac{S}{4}, \frac{3 gS}{2}$
(D) $\frac{S}{4}, \frac{3 gS}{2}$ ✓ Correct

Answer: (D)

Solution

Let the particle be at height

x

above the ground after falling through

(S - x)

, so

v^{2} = 2 g (S - x)

. Its potential energy is

m g x

and its kinetic energy is

\frac{1}{2} m v^{2}

. The condition

KE = 3 PE

gives

\frac{1}{2} v^{2} = 3 g x \Rightarrow v^{2} = 6 g x .

Equating the two expressions for

v^{2}

2 g (S - x) = 6 g x

, so

S = 4 x

and

x = \frac{S}{4}

. The speed there is

v = 2 g (S - \frac{S}{4}) = \frac{3 g S}{2} .

The height is

\frac{S}{4}

and the speed is

\frac{3 g S}{2}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Work, Power & Energy

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