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

CONCEPT:

According to the third equation of motion we have initial velocity u, final velocity v, a is the acceleration, and s is the distance. It is written as;

v2 = u2 + 2as

CALCULATIONS:

Let us suppose that the height of the body is y.

When the body falls from the rest its height will be (S-y), which is shown in the figure below:

Using three equations of motion we have;

v2 = 02 + 2g(S - y) ---(1)

here, g is the acceleration due to the gravity of the failing body, and S-y is the distance in which the body falls.

Further solving the equation (1) we get;

 ----(2)

When the body is in the height y from the ground, its potential energy (U) is written as;

U =mgy ---(3)

It is given that the kinetic energy (K) is three times its potential energy (U), therefore according to the condition we have,

K=3U ----(4)

where,  

Putting the values of kinetic energy and potential energy in equation (4) we get;

Using equation (1) we get;

Further solving we get;

S - y = 3y

⇒ ----(5)

Now, by putting the value of y in equation (2) we get;

⇒

⇒

Hence, option 1) is the correct answer.