A particle is fired vertically from the surface of the earth with a velocity `kupsilon_(e)`, where `upsilon_(e)` is the escape velocity and `klt1`. Neglecting air resistance and assuming earth's radius as `R_(e)`. Calculated the height to which it will rise from the surface of the earth.

Answer

`(R_(e )k^(2))/(1-k^(2))`

Explanation

We know `v_(c) = sqrt((2GM)/(R_(e )) rArr GM (R_(e )v_(e )^(2))/(2)` ….(i)
From energy conservation
`U_(i)+K_(i)=U_(r )+K_(f)`
`-(GMm)/(R_(e ))+(1)/(2)m(kv_(e ))^(2) = -(GMm)/((R_(e )+h))+0 rArr - (k^(2)v_(e )^(2))/(2)=(-GM)/(R_(e )+h)` ....(ii)
From (i) and (ii) `-(v_(e )^(2))/(2)+(k^(2)v_(e )^(2))/(2)=-(R_(e )v_(e )^(2))/(2(R_(e )+h))rArr 1-k^(2)=(R_(e ))/((R_(e )+h))`
`(R_(e )+h)/(R_(e ))=(1)/(1-k^(2)) rArr h = (k^(2)R_(e ))/(1-k^(2))`