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Q. 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

AP EAMCETAP EAMCET 2018

Solution:

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