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

Explanation

At the surface of the earth, potentila energy of theearth partile system is `-(Gmm)/R` with usual symbols. The kinetilc energy is `1/2mv_0^2` where `v=)=9.8mkms^-1`. At the maimu height the kinetic energy is zro. If the maximum height reached is H, the potential energyof the earth particle system at this instant is `-(GMm)/(R+H)`. Usig conservation of energy.
`-(Gmm)R+1/2mv_0^2=-(GMm)/(R+H)`
Writing `GM=gR^2` and divindin by m
`-gR+v_0^2/2=(-gR^2)/(R+H)`
or, R^2/(R+H)=R-(v_0^2)/(2g)`
or ` (R+H)=R^2/(R-v_0^2)/(2g)`
putting the values of `R, v_0 and g` on the right side,
`R+H=((6400km)^2)/(6400km-((9.8kms^-2)^2)/(2xx9.8ms^-2))`
or H=(2730-6400)kmk=20900km`.