A body of `200 g` begins to fall from a height where its potential energy is `80 J`. Its velocity at a point where its kinetic and potential energies are equal `(in m//s)`.

Options

A. `10 sqrt(8)`
B. 4
C. 400
D. 20

Answer

D

Explanation

`T.E` at top = `T.E` at given point , `P.E = P.E^(1) + K.E^(1)`
`80=2K.E^(1)(because P.E^(1) = K.E^(1))`.