A particle is released from a height $H$. At a certain height, its kinetic energy is half of its potential…

A particle is released from a height $H$. At a certain height, its kinetic energy is half of its potential energy with reference to the surface of the earth. Height and speed of the particle at that instant are respectively
  1. $\frac{H}{3}, \sqrt{\frac{2 g H}{3}}$
  2. $\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}$
  3. $\frac{2 H}{3}, \sqrt{2 g H}$
  4. $\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}$

Solution

Total mechanical energy of particle, i.e.
Given, $ \mathrm{KE}=\frac{1}{2} \mathrm{PE} $ i.e. $ \begin{aligned} & \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{1}{2} \\ & \mathrm{PE}=2 \mathrm{KE} \end{aligned} $ or Substituting in this Eq. (i), we get $ \begin{aligned} \mathrm{PE}+\mathrm{KE} & =m g H \\ 2 \mathrm{KE}+\mathrm{KE} & =m g H \\ 3 \mathrm{KE} & =m g H \\ \mathrm{KE} & =\frac{m g H}{3} \\ \mathrm{PE} & =\frac{2}{3} m g H \end{aligned} $ Similarly, So, height from ground at that instant, $ h=\frac{2 H}{3} $ So, speed of particle, $ v=\sqrt{2 g h}=\sqrt{2 g H / 3} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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