An object is projected vertically upwards with a velocity u from the surface of the earth. Show that the maximum height reached by the object is `h=(u^2R)/(2gR-u^2)`
where R is the radius of the earth. Calculate escape velocity from it.

Explanation

Let m be the mass of the projected object.
`therefore` Initial kinetic energy `=1/2mu^2`
and potential energy at the surface,
`U=-(GMm)/((R+h))`
According to the law of conservation of energy , total energy at the earth.s surface =total energy at height h
`therefore 1/2 mu^2+(-(GMm)/R)=-(GMm)/(R+h)+0`
or, `1/2 mu^2=GMm[1/R-1/(R+h)]`
or,`u^2=(2GMh)/(R(R+h))[because GM=gR^2]`
or, `u^2(R+h)=2gRh or, h(2gR-u^2)=u^2R`
or,`h=(u^2R)/(2gR-u^2)` [Proved] .........(1)
From equations (1) ,
`2gR-u^2=(u^2R)/(h)`
If the object starts with escape velocity then `h to oo`
or, `(v_e^2R)/hto0`
`therefore 2gR-v_e^2=0 therefore v_e=sqrt(2gR)`