Question

At what height above the ground the kinetic energy and potential energy of a body falling freely from rest from a height H are equal (neglect air resistance)

Answer 1

Answer:

H/2

Explanation:

at x above the surface of ground

kinetic energy = potential energy

1/2 mv² = mgx

1/2 m (u²+2g[H-x]) =mgx

here initial velocity,u = 0

1/2 m [2g(H-x)] =mgx

mgH - mgx =mgx

mgH =2mgx

x =H/2