A particle is projected vertically with speed V from the surface of the earth . Maximum height attained by the particle , in term of the radius of earth R,V and g is ( V `lt` escape velocity , g is the acceleration due to gravity on the surface of the earth )

Explanation

At the maximum height kinetic energy of the particle is zero. So, using conservation of mechanical energy , `1/2 mv_(0)^(2) - (GMm)/R = (-GMm)/(R+H)`
or , `1/2 mv_(0)^(2) = GMm (1/R - 1/(R+H)) = (GMmH)/(R(R+H))`
or , `v_(0)^(2) = (2gH)/(1+H/R)`
simplifying for H , we get `H = (v_(0)^(2)R)/(2gR-v_(0)^(2))`
(ii) Increase in gravitational potential energy is
`U_(h)-U_(0) = (-GMm)/(R+h) = (-GMm)/R`
` = GMm(1/R -1/(R+h))=(mgh)/(1+h//R) " " ` [ as in (i)]
Note : If ` h lt lt R , h/R ` is negligible , hence `U_(h)-U_(0)` = mgh
If `U_(0) = 0 ` (assumed ) , then `U_(h) = mgh `