Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$\frac{S}{2}, \frac{\sqrt{3gS}}{2}$

$\frac{S}{4}, \sqrt{\frac{3gS}{2}}$

$\frac{S}{4}, \frac{3gS}{2}$

$\frac{S}{4}, \frac{\sqrt{3gS}}{3}$

Step-by-Step Solution

Conservation of Mechanical Energy: The total mechanical energy ( $E$ ) of the particle is conserved throughout the motion. At the release point (height $S$ ), the velocity is zero. $E = K_i + U_i = 0 + mgS = mgS$ (Assuming potential energy $U=0$ at the surface).
Condition at Specific Height: Let the particle be at height $h$ and have speed $v$ . The problem states that Kinetic Energy ( $K$ ) is three times Potential Energy ( $U$ ). $K = 3U$
Calculate Height ( $h$ ): $E = K + U = 3U + U = 4U$ Substituting the expressions: $mgS = 4(mgh) \implies h = \frac{S}{4}$
Calculate Speed ( $v$ ): The Potential Energy at this height is $U = mg\left(\frac{S}{4}\right)$ . The Kinetic Energy is $K = 3U = \frac{3mgS}{4}$ . Using the formula $K = \frac{1}{2}mv^2$ : $\frac{1}{2}mv^2 = \frac{3mgS}{4}$ $v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{2}}$

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