Aprojectile is shot straight up from the earth's surface at a speed of 10,000 km/hr. how high does it go?

Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius. 
KE/m = (9000000/3600)^2/2 = 3858025 J/kg 
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r 
KE/m = ΔPE/m 
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m 
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)