Find the minimum initial speed of a projectile in order for it to reach a height of 2222 km above the surface of the earth.

Answer 1

iniial KE=final PE
1/2 m v^2=mgh
solve for velocity v.

Answer 2

V^2 = Vo^2 + 2g*h = 0.
Vo^2 + (-19.6)2.222*10^6 = 0,
Vo^2 = 43.55*10^6,
Vo = 6,599.3 m/s.