CBSE Class 9 Answered

At position A:

Initial velocity of body = 0.

Thus, kinetic energy = K =0

Potential energy = U = mgh

Hence, total energy = K + U = mgh

At position B:

Let v₁ be the velocity acquired by the body after falling through distance x.

Then , u = 0, S = x, a = g

From equation v² = u² + 2aS, we have

 V₁²  = 0 + 2gx = 2gx

Thus, kinetic energy is

K = 1/2 m V₁²

   = 1/2 x m x 2gx 

   = mgx

Thus, potential energy is = mg (h - x)

Hence, total energy = K + U = mgx + mg(h -x) = mgh

At position C:

Let v be the velocity acquired by the body on reaching the ground.

Then , u = 0, S = h, a = g

From equation v² = u² + 2aS, we have

 V₁²  = 0 + 2gh = 2gh

Thus, kinetic energy is,

K = 1/2 m V₁²

   = 1/2 x m x 2gh

Now at C, height of body above the ground = 0

Thus, potential energy is 0.

Hence, total energy = K + U = mgh + 0 = mgh