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

Answer

D

Explanation

We can realise the situation as shown. Let at point A, the particle.s kinetic energy is three times its potential energy.

Velocity at C,
`v^(2)=0+2gx`
`v^(2)=2gx …(i)`
Potential energy at C
`=mg(S-x) ...(ii)`
At point C,
kinetic energy = `3xx` potential energy
i.e., `(1)/(2)mxx2gx=3xxmg(S-x)`
or `x=3S-3x`
or `4x=3S`
or `S=(4)/(3)x`
or `x=(3)/(4)S`
Therefore, from Eq. (i)
`v^(2)=2gxx(3)/(4)S`
or `v^(2)=(3)/(2)gS`
or `v=sqrt((3)/(2)gS)`
Height of the particle from the ground
`=S-x`
`S-(3)/(4)S=(S)/(4)`