Super Missile

The most important part of the question is that the velocity of missile at highest point will not be zero . So let the final velocity of missile be $v^{\prime}$ which will be perpendicular to the line joining the missile and center of earth.

From conservation angular momentum along the line perpendicular to line joining missile and center of earth-

$mR_e \times (v\sin{60^{\circ}}) = mv^{\prime}\times (R_e + h) \quad$ , where $h$ represents the maximum height reached by the missile.

$\therefore v^{\prime} = \displaystyle \dfrac{\sqrt{3}R_e v}{2(R_e + h)}$

From conservation of energy-

$\displaystyle \dfrac 12\times mv^2 - \dfrac{GmM_e}{R_e} = -\dfrac{GmM_e}{R_e+h} + \dfrac 12\times m(v^{\prime})^2$

Now plug in the values of $v^{\prime}$ and $v$ and you get a quadratic equation which on solving gives $(R_e + h)= 1.235R_e$

$\therefore h = \boxed{0.23R_e}$