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.

An object launched into motion whose subsequent trajectory is determined by external forces. The ideal elementary projectile motion model retains uniform gravity and neglects air resistance.

The horizontal displacement between launch and landing in a projectile motion model. With equal launch and landing heights, uniform gravitational acceleration $g$, no air resistance, launch speed $u$ and angle $\theta$, it is $R=u^2\sin (2\theta )/g$. Different landing heights or forces change the result.

Under uniform gravity without drag, a projectile launched at speed $u$ from height $h$ reaches a maximum horizontal projectile range $u\sqrt{u^2+2gh}/g$. For horizontal position $X$, maximize $y=h+Xz-g{X}^2(1+z^2)/(2u^2)$ over $z=\tan \alpha$. The maximizer is $z=u^2/(gX)$ and its envelope meets the ground at the displayed distance. The optimum angle obeys $\tan \alpha =u/\sqrt{u^2+2gh}$.

A projectile launched from and returning to the floor beneath a horizontal ceiling of height $H$ has maximum projectile range $\sqrt{8H(u^2/g-2H)}$ when $0\le H\le u^2/(4g)$. For higher ceilings the maximum is $u^2/g$. This assumes uniform gravitational acceleration, no air resistance and fixed launch speed $u$.

Indeed, with $z={\sin }^2\theta$, the apex condition is $z\le 2gH/u^2$ and $R^2=4u^{4}z(1-z)/g^2$. This concave quadratic function is maximized at the admissible value nearest $1/2$. Thus the lower-ceiling expression must not be extrapolated past its regime. If touching the ceiling is excluded, the limiting value is a supremum instead of an attained maximum.

For a jet arriving horizontally at a target displaced horizontally by w and upward by h, the target lies at the projectile apex. Vertical kinematics gives $u_y^2=2gh$ and flight time $u_y/g$; the horizontal travel then fixes the displayed $u_x$. The launch speed is $\sqrt{u_x^2+u_y^2}$ and $\tan \theta =2h/w$. The ideal ballistic approximation neglects air drag, breakup and internal jet stresses after launch.

Steady incompressible mass flow rate stays constant along a jet, while speed and cross-sectional area can change under gravity. At a horizontal apex the arriving speed is the initial horizontal component. The force on a target that stops this horizontal momentum is $\dot{m}{v}_{\mathrm{hit}}$, using the conserved nozzle mass flux. Substituting the nozzle area into $\rho A_{\mathrm{hit}}{v}_{\mathrm{hit}}^2$ without adjusting the area gives the wrong result.

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.