A projectile 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)