Projectile motion

Motion after launch under a specified gravitational field, with negligible drag force and no thrust. In a constant downward gravitational acceleration $g$, horizontal velocity is constant and vertical position obeys $y(t)=y_0+v_{y0}t-g{t}^2/2$. Eliminating time gives a parabola when horizontal velocity is nonzero. Over larger distances the changing direction and magnitude of gravity require an orbital model instead.

In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed $v_o$, its earlier velocity at angle $\theta$ above the horizontal has vertical component $v_o\tan \theta$. Its remaining height gain is therefore $v_o^2{\tan }^2\theta /(2g)$. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing $v_o=\sqrt{g(R_E+h_o)}$ assumes constant gravitational magnitude also at the orbit; computing $v_o=\sqrt{g_0{R}_E^2/(R_E+h_o)}$ instead assumes an inverse-square law with surface value $g_0$. Those assumptions must be distinguished.