A particle is released from height S from the surface of the earth. At a certain height its KE is three times its PE. The height from the surface of earth and the speed of the particle at that instant are resp.

Options

A. `S/4.(sqrt(3gS))/(2)`
B. `S/2.(sqrt(3gS))/(2)`
C. `S/4.sqrt((3gS)/(2))`
D. `S/4.(3gS)/(2)`

Answer

C

Explanation

`mgS=mgS.+1/2mv^2` ….(i)
`1/2mv^2=3(mgS.)` ….(ii)
from equation (i) and (ii), we get
`mgS=mgS.+3mgS.`
`S.=S/4`
`1/2mv^2=3mgS/4`
`v=sqrt((3gS)/2)`