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

Answer

D

Explanation

Velocity of B –
`v^2 = 2(S - x)g`
`KE = 3 xx PE`
`=1/2 m xx 2(S-x) g = 3 xx mgx`
`implies S - x = 3x implies x =S/4`
`v^2 =2g(S-x)=2g(S-S/4)=(2g(3))/(4)=(3Sg)/2`
`v=sqrt((3gS)/(2))`