A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively 

Options

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

Answer

D

Explanation

Velocity at B when dropped from A where AC = S.
`v^2 = 0 + 2g(S - x)" "..(i)`
or `v^2 = 2g(S - x)`
potential energy at `B = mgx`
`:. ` Kinetic energy = `3 xx "potential energy"`
`:. 1/2 m xx 2 g (S - x) = 3 xx mgx`
or `S - x = 3x " or " S = 4x " or " x = S/4`
Putting this value of x in eqn. (i), we get
`v^2 = 2g (S - S/4) = (2g xx 3S)/(4) = (3gS)/(2) " or " v = sqrt((3gS)/(2))`
Hence, `x = S/4` and `v = sqrt((3gS)/(2))`