Answer
$5.644\space km/s$
Work Step by Step
Here we use the principle of conservation of mechanical energy.
Final mechanical energy = Initial mechanical energy
$K+U=K_{0}+U_{0}$
$-\frac{GM_{E}m}{r}=\frac{1}{2}mV_{0}^{2}-\frac{GM_{E}m}{R_{E}}$
Where m is the rocket's mass, r is the distance from Earth's center at the peak & Earth's radius $R_{E}$ is the distance at launch. Solving for r gives.
$\frac{GM_{E}}{R_{E}}-\frac{1}{2}V_{0}^{2}=\frac{GM_{E}}{r}$
$2GM_{E}(\frac{1}{R_{E}}-\frac{1}{r})=V_{0}^{2} $; Let's plug known values into this equation.
$V_{0}^{2}=2\times6.67\times10^{-11}Nm^{2}/
kg^{2}\times5.97\times10^{24}kg(\frac{1}{6.37\times10^{6}m}-\frac{1}{8.52\times10^{6}m})$
$V_{0}^{2}=79.64\times10^{7}(0.16-0.12)\space m^{2}/s^{2}$
$V_{0}=5644.1\space m/s= 5.644\space km/s$
