A particle is projected vertivally upwards from the surface of earth `(radius R_e)` with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is ……

Explanation

Let `upsilon_(e)` be the escape velocity of Earth. A particle can escape when its total energy `(KE + PE)` becomes zero at infinity. So
`(1)/(2) m upsilon_(e)^(2) + (-(GMm)/(R )) = 0` or `(1)/(2) m upsilon_(e)^(2) = (GMm)/(R )`
Supplied `KE = (1)/(2) xx (1)/(2) m upsilon_(e)^(2) = (GMm)/(2R)`
Let the particle rises to a height `h`, then
`(1)/(2) xx (1)/(2) m upsilon_(e)^(2) = (GMm)(R + h)`
`(GMm)/(2R) = (GMm)/(R + h)` or `h = R`