A particle is released from height S above th | vedclass

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(D) Let the height from the surface be $x$. The distance fallen by the particle is $(S - x)$.Using the equation of motion $v^2 = u^2 + 2as$,where $u =

Vedclass · Accepted Answer

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

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(D) Let the height from the surface be $x$. The distance fallen by the particle is $(S - x)$.Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$ and $a = g$,we get $v^2 = 2g(S - x)$.The potential energy at height $x$ is $PE = mgx$.The kinetic energy at that instant is $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(S - x)) = mg(S - x)$.According to the problem,$KE = 3 	imes PE$.Substituting the expressions: $mg(S - x) = 3(mgx)$.Dividing both sides by $mg$: $S - x = 3x$.$S = 4x \Rightarrow x = \frac{S}{4}$.Now,substitute $x$ into the velocity equation: $v^2 = 2g(S - \frac{S}{4}) = 2g(\frac{3S}{4}) = \frac{3gS}{2}$.Therefore,$v = \sqrt{\frac{3gS}{2}}$.The height is $\frac{S}{4}$ and the speed is $\sqrt{\frac{3gS}{2}}$.

$A$ particle is released from 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 surface of the earth and the speed of the particle at that instant are respectively:

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