A body of mass `2 kg` is thrown up vertically with kinetic energy of `490 J`. If `g = 9.8m//s^(2)`, the height at which the kinetic energy of the body becomes half of the original value, is

Options

A. `50m`
B. `25m`
C. `12.5 m`
D. `19.6 m`

Answer

C

Explanation

At projection,`T.E = K.E = 490`
at given point, `P.E^(1)+K.E^(1) = T.E^(1)`
`mgh+(K.E)/(2)=T.E`