Answer
The projectile reaches a radial distance of $~~\frac{4}{3}~R_E$
Work Step by Step
Let $M_E$ be the mass of the Earth.
The escape speed is $v_{esc} = \sqrt{\frac{2GM_E}{R_E}}$
We can find the speed $v$ if $v$ is half the escape speed:
$v = \frac{1}{2}\cdot \sqrt{\frac{2GM_E}{R_E}}$
$v = \sqrt{\frac{GM_E}{2R_E}}$
Let $m$ be the mass of the projectile.
We can use conservation of energy to find the radial distance $r$ that a projectile reaches:
$K_2+U_2 = K_1+U_1$
$0-\frac{GM_E~m}{r} = \frac{1}{2}mv^2-\frac{GM_E~m}{R_E}$
$-\frac{GM_E~m}{r} = \frac{1}{2}m(\sqrt{\frac{GM_E}{2R_E}})^2-\frac{GM_E~m}{R_E}$
$-\frac{GM_E~m}{r} = \frac{1}{2}m(\frac{GM_E}{2R_E})-\frac{GM_E~m}{R_E}$
$-\frac{GM_E~m}{r} = \frac{GM_E~m}{4~R_E}-\frac{GM_E~m}{R_E}$
$-\frac{1}{r} = \frac{1}{4~R_E}-\frac{1}{R_E}$
$-\frac{1}{r} = \frac{1}{4~R_E}-\frac{4}{4~R_E}$
$-\frac{1}{r} = -\frac{3}{4~R_E}$
$r = \frac{4}{3}~R_E$
The projectile reaches a radial distance of $~~\frac{4}{3}~R_E$
