#### Answer

The projectile goes up to a height of 420 km.

#### Work Step by Step

We can convert the initial speed to units of m/s.
$v = (10,000~km/h)(1000~m/km)(1 ~hr/3600~s)$
$v = 2777.8~m/s$
We can use conservation of energy to find the height $h$ of the projectile when the speed decreases to zero. Let $R$ be the earth's radius. Let $M_e$ be the earth's mass. Let $M_p$ be the projectile's mass.
$K_f+U_f = K_0+U_0$
$0+U_f = K_0+U_0$
$-\frac{G~M_e~M_p}{R+h} = \frac{1}{2}M_pv^2-\frac{G~M_e~M_p}{R}$
$-2~G~M_e~R = (R)(R+h)(v^2)-(2~G~M_e)(R+h)$
$2~G~M_e~h-R~h~v^2 = R^2~v^2$
$h = \frac{R^2~v^2}{2~G~M_e-R~v^2}$
$h = \frac{(6.38\times 10^6~m)^2(2777.8~m/s)^2}{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)-(6.38\times 10^6~m)(2777.8~m/s)^2}$
$h = 4.20\times 10^5~m$
$h = 420~km$
The projectile goes up to a height of 420 km.