Answer
The projectile reaches a radial distance of $~~2~R_E$
Work Step by Step
Let $M_E$ be the mass of the Earth.
Let $m$ be the mass of the projectile.
In order to escape the Earth:
$K+U = 0$
$K = -U$
$K = \frac{G~M_E~m}{R_E}$
If the kinetic energy is half of the required kinetic energy:
$K_1 = \frac{G~M_E~m}{2~R_E}$
We can use conservation of energy to find the radial distance $r$ that a projectile reaches:
$K_2+U_2 = K_1+U_1$
$0-\frac{GM_E~m}{r} = \frac{G~M_E~m}{2~R_E}-\frac{GM_E~m}{R_E}$
$-\frac{1}{r} = \frac{1}{2~R_E}-\frac{1}{R_E}$
$-\frac{1}{r} = \frac{1}{2~R_E}-\frac{2}{2~R_E}$
$-\frac{1}{r} = -\frac{1}{2~R_E}$
$r = 2~R_E$
The projectile reaches a radial distance of $~~2~R_E$
