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.

Options

A. `(H)/(3), sqrt((2gH)/(3))`
B. `(H)/(3), 2sqrt((gH)/(3))`
C. `(2H)/(3).sqrt(2gH)`
D. `(2H)/(3).sqrt((2gH)/(3))`

Answer

D

Explanation

Total mechanical energy of particle i.e.,
PE + KE = mgH
Given, `KE = (1)/(2)PE`
i.e., `(KE)/(PE) = (1)/(2)`
or PW = 2KE
Substituting in this Eq. (i) , we get
PE + KE = mgH
`2KE + KE = mgH`
3KE = mgH
`KE = (mgH)/(3)`
Similarly, `PE = (2)/(3) mgH`
So, height from ground at that instant,
`h = (2H)/(3)`
So, speed of particle,
`v = sqrt(2gh) = sqrt(2gH//3)`