Find the minimum initial speed of a projectile in order for it to reach a height of 2022 km above the surface of the earth.

Answer 1

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The formula I found is
v=sqrt(2gY), where Y is the maximum height and g is -9.8m/s^2
Obviously the units would be sqrt ( m * m/s^2) = m/s, so it matches with velocity.
We convert 2022km to m, then plug into formula to get v = 4451.47167 m/s

Makes sense -- the projectile has to go mad fast to go that far into space