A projectile is fired from the Earth's surface with a speed v₀ = √gR = 8.0 km/s, in a direction 30°…

Step 1. Angular momentum. At launch the velocity makes $30^\circ$ with the radius, so its part at right angles to the radius is $v_0\sin 30^\circ$. At the highest point, distance $r$ from the centre, the velocity $v$ is at right angles to the radius: $v_0 R\sin 30^\circ = vr$, so $v = \dfrac{v_0 R}{2r}$.

Step 2. Energy (per kilogram). At launch: $\tfrac12 v_0^2 - \dfrac{GM}{R} = \tfrac12 gR - gR = -\tfrac12 gR$. At the top: $\tfrac12 v^2 - \dfrac{gR^2}{r}$. Put in $v$ from Step 1:

Multiply by $\dfrac{8r^2}{gR}$: $4r^2 - 8Rr + R^2 = 0$, so $r = R\left(1 \pm \dfrac{\sqrt3}{2}\right)$.

Step 3. Height. The larger root is the highest point: $r = 1.866R$, so the height is $\dfrac{\sqrt3}{2}R = 5543\ \text{km}$. (The smaller root would be the nearest point of the orbit, inside the Earth.)

Step 4. Speed at the top. $v = \dfrac{v_0}{2 \times 1.866} = \dfrac{8000\ \text{m/s}}{1.866 \times 2} = 2144\ \text{m/s}$.

Step 5. With constant $g$. The vertical part of the launch velocity is $v_0\cos 30^\circ = 6928\ \text{m/s}$, and $\dfrac{(6928\ \text{m/s})^2}{2 \times 10\ \text{m/s}^2} = 2400\ \text{km}$, less than half of the true height. Gravity weakens with height, so the projectile climbs much higher.