Question

a body has 500 joules of potential energy on top of the building now its start falling so what will its kinetic energy at half of the building height

Answer 1

(SEE THE ATTACHMENT FOR FIGURE)

Consider the situation that the ball is dropped from the building,,

At the top of the building,

P.E = mgh = 500J

At point B which is the half of the height of the building,

So, the total energy at B is

P.E + K.E

mgx + 1/2mv^2 ----------- (i)

from 3rd equation of motion,

v^2 = 2gh

here, v^2 = 2g(h-x)

so, from equation (i),

T.E = mgx + 1/2m2g(h-x)

= mgx + mg(h-x)

= mgx + mgh - mgx

T.E = mgh = 500J

so, K.E = mgh - mgx

,. = mgh - mg(h/2)

= mg( h - h/2)

= mg(h/2)

= mgh/2 = 500J/2 = 250J.