A rocket starts vertically upward with speed `v_(0)`. Show that its speed `v` at height `h` is given by `v_(0)^(2)-v^(2)=(2hg)/(1+h/R)`
where `R` is the radius of the earth and `g` is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed `0.9` times the escape velocity.

Answer

4.26R

Explanation

`v(dv)/(dh)=-(GM)/((R+h)^(2)),int_(v_(0))^(v)vdv=-int(dR^(2))/((R+h)^(2))dh`
`(v^(2)-v_(0)^(2))/(2)=(-gR^(2))[(-1)/(R+h)]_(0)^(h)`
`(v-v_(0)^(2))/(2)=+gR^(2)[(1)/(R+h)-(1)/(R)]`
`v_(0)^(2)-v^(2)=(2gRh)/((R+h)),v_(0)^(2)-v^(2)=(2gh)/((1+(h)/(R)))`