With what velocity must a body be thrown from earth's surface so that it may reach a maximum height of `4R_(e)` above the Earth's surface ? (Radius of the Earth `R_(e)=6400 km,g=9.8 ms^(-2)`).

Explanation

By using conservation of mechanical energy `1/2 m_(0) v^(2) - (GMm_(0))/R_(e) =0- (GMm_(0))/((R_(e)+4R_(e)))`
`1/2 m_(0)V^(2)=-(GMm_(0))/(5R_(e))+(GMm_(0))/R_(e) implies 1/2 m_(0)v^(2)=4/5 (GMm_(0))/R_(e)implies v^(2)=8/5 (GM)/R_(e)=8/5 (gR_(e)^(2))/R_(e)`
`v^(2)=8/5xx9.8xx6400xx10^(3)=10^(8) implies v=10` km/s.