If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is three-fifths the escape speed?

Question

I keep getting the answer 2.56x10^6 m after solving it a few different ways. Help would be great.

Answer 1

Its energy at launch is PE at surface of earth.

1/2 mv^2=GMe*M/Re

so when is GPE= (2/5)^2*GMeM/Re ?

or GMeM/(Re+H)=4/25*GMeM/Re

4(Re+h)=25Re
or 4h=21Re
h= 21Re/4 or near 5.25Re
or h= about 6.5*5.2E6 meters which is not your answer. check my work.