A particle is fired verticaly upward with a speed of 9.8 km`s^-1`. Find the maximum height attained by the particle. Radius of earth =6400 k,m and g at the surface`=9.8ms^2`. Consider only earth's gravitation.

Explanation

Initial energy of particle on earth's surface is
`E_(r ) = (1)/(2)m u^(2) - (GM m)/(R )`
If the particle reaches upto a height h above the surface of earth then its final energy will only be the gravitational potential energy.
`E_(f) = -(GM m)/(R + h)`
According to energy conservation, we have
`E_(t) = E_(f)`
or `(1)/(2)m u^(2) - (GM m)/(R ) = -(GM m)/(R+h)`
or `(1)/(2)u^(2) - gR = -(gR^(2))/(R + h)`
or `h = (2gR^(2))/(2gR - u^(2)) - R`
`= (2 xx 9.8 xx (6400 xx 10^(3))^(2))/(2 x 9.8 xx 6400 xx 10^(3) - (9.8)^(2)) - 6400 xx 10^(3)`
`= (27300 - 6400) xx 10^(3) = (20900) km`