拋體運動者，物理公式也，述拋一物之（二維）軌跡。其軌跡名拋物線也。
按運動方程，設 x {\displaystyle x} ， y {\displaystyle y} 二軸；又設其始速 v 0 {\displaystyle v_{0}} ，交水平之角（自 x {\displaystyle x} 軸逆時升之角） θ {\displaystyle \theta } ，無氣阻則：
v x = v 0 cos θ {\displaystyle v_{x}=v_{0}\cos \theta }
v y = v 0 sin θ − g t {\displaystyle v_{y}=v_{0}\sin \theta -gt}
s x = v 0 cos θ t {\displaystyle s_{x}=v_{0}\cos \theta t}
s y = v 0 sin θ t − 1 2 g t 2 {\displaystyle s_{y}=v_{0}\sin \theta t-{\frac {1}{2}}gt^{2}}
然此未足。需計飛之時、高、水平距，兼系 x {\displaystyle x} ， y {\displaystyle y} 座標之方程。下又設首尾垂直位移者 d {\displaystyle d} 。
d = v 0 sin θ t − 1 2 g t 2 {\displaystyle d=v_{0}\sin \theta t-{\frac {1}{2}}gt^{2}}
1 2 g t 2 − v 0 sin θ t + d = 0 {\displaystyle {\frac {1}{2}}gt^{2}-v_{0}\sin \theta t+d=0}
t = v 0 sin θ ± v 0 2 sin 2 θ − 2 g d g {\displaystyle t={\frac {v_{0}\sin \theta \pm {\sqrt {v_{0}^{2}\sin ^{2}\theta -2gd}}}{g}}}
然若 d < 0 {\displaystyle d<0} ，則 v 0 sin θ − v 0 2 sin 2 θ − 2 g d g {\displaystyle {\frac {v_{0}\sin \theta -{\sqrt {v_{0}^{2}\sin ^{2}\theta -2gd}}}{g}}} 者負數也。故：
t = { v 0 sin θ ± v 0 2 sin 2 θ − 2 g d g , h ≥ 0 v 0 sin θ + v 0 2 sin 2 θ − 2 g d g , h < 0 {\displaystyle t={\begin{cases}{\frac {v_{0}\sin \theta \pm {\sqrt {v_{0}^{2}\sin ^{2}\theta -2gd}}}{g}},h\geq 0\\{\frac {v_{0}\sin \theta +{\sqrt {v_{0}^{2}\sin ^{2}\theta -2gd}}}{g}},h<0\end{cases}}}
0 = ( v 0 sin θ − g ( 0 ) ) 2 − 2 g h {\displaystyle 0=(v_{0}\sin \theta -g(0))^{2}-2gh}
h = v 0 2 sin 2 θ 2 g {\displaystyle h={\frac {v_{0}^{2}\sin ^{2}\theta }{2g}}}
R {\displaystyle R}
= v 0 cos θ t {\displaystyle =v_{0}\cos \theta t}
= v 0 cos θ ( v 0 sin θ ± v 0 2 sin 2 θ − 2 g d g ) {\displaystyle =v_{0}\cos \theta ({\frac {v_{0}\sin \theta \pm {\sqrt {v_{0}^{2}\sin ^{2}\theta -2gd}}}{g}})}
= { v 0 2 sin 2 θ ± v 0 v 0 2 sin 2 2 θ − 8 g d cos 2 θ 2 g , d ≥ 0 v 0 2 sin 2 θ − v 0 v 0 2 sin 2 2 θ − 8 g d cos 2 θ 2 g , d < 0 {\displaystyle ={\begin{cases}{\frac {v_{0}^{2}\sin 2\theta \pm v_{0}{\sqrt {v_{0}^{2}\sin ^{2}2\theta -8gd\cos ^{2}\theta }}}{2g}},d\geq 0\\{\frac {v_{0}^{2}\sin 2\theta -v_{0}{\sqrt {v_{0}^{2}\sin ^{2}2\theta -8gd\cos ^{2}\theta }}}{2g}},d<0\end{cases}}}
t = s x v 0 cos θ {\displaystyle t={\frac {s_{x}}{v_{0}\cos \theta }}}
s y {\displaystyle s_{y}}
= v 0 sin θ ( s x v 0 cos θ ) − 1 2 g ( s x v 0 cos θ ) 2 {\displaystyle =v_{0}\sin \theta ({\frac {s_{x}}{v_{0}\cos \theta }})-{\frac {1}{2}}g({\frac {s_{x}}{v_{0}\cos \theta }})^{2}}
= s x tan θ − g s x 2 2 v 0 2 cos 2 θ {\displaystyle =s_{x}\tan \theta -{\frac {gs_{x}^{2}}{2v_{0}^{2}\cos ^{2}\theta }}}
即：
y = x tan θ − g x 2 2 v 0 2 cos 2 θ {\displaystyle y=x\tan \theta -{\frac {gx^{2}}{2v_{0}^{2}\cos ^{2}\theta }}}
若 d = 0 {\displaystyle d=0} ，則：
t {\displaystyle t}
= v 0 sin θ ± v 0 2 sin 2 θ − 2 g ( 0 ) g {\displaystyle ={\frac {v_{0}\sin \theta \pm {\sqrt {v_{0}^{2}\sin ^{2}\theta -2g(0)}}}{g}}}
= 2 v 0 sin θ g {\displaystyle ={\frac {2v_{0}\sin \theta }{g}}}
= v 0 2 sin 2 θ ± v 0 v 0 2 sin 2 2 θ − 8 g ( 0 ) cos 2 θ 2 g {\displaystyle ={\frac {v_{0}^{2}\sin 2\theta \pm v_{0}{\sqrt {v_{0}^{2}\sin ^{2}2\theta -8g(0)\cos ^{2}\theta }}}{2g}}}
= v 0 2 sin 2 θ g {\displaystyle ={\frac {v_{0}^{2}\sin 2\theta }{g}}}
，
二軸；又設其始速
，交水平之角（自 軸逆時升之角）
，無氣阻則：
， 座標之方程。下又設首尾垂直位移者 
。
，則
者負數也。故：
之殊況 ，則：
