自由落体抛物线，与 v 矢量相切

Question

我有照片，

请帮我画一条从点 A 切向 V0 的抛物线。

 \begin{tikzpicture}
 \draw  (0,3) -- (3,3)-- node[left] {H} (3,2) -- node[below] {$L_{max}$} (6,2);
 \draw [dashed]   (3,3) -- +(0:1.5);
 \draw [->] [>=stealth,line width=0.3mm]   (3,3) -- +(30:1.5) node [near end, above] {${\vec{v}}_0$};
\draw (4,3) arc (0:30:1) node at (38:5.3) {$\alpha$};
 \end{tikzpicture}

谢谢你的帮助

Answer 1

你可以使用类似如下的近似值

\draw [red,thick,dashed] (3,3) to[out=30,in=100] (5.5,2);

in=100表示路径从 100 度方向到达其端点。请根据您的喜好进行调整。

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
 \draw  (0,3) -- (3,3)-- node[left] {H} (3,2) -- node[below] {$L_{\mathrm{max}}$} (6,2);
 \draw [dashed]   (3,3) -- +(0:1.5);
 \draw [->,>=stealth,line width=0.3mm]   (3,3) -- +(30:1.5) node [near end, above] {${\vec{v}}_0$};
 \draw (4,3) arc (0:30:1) node at (38:5.3) {$\alpha$};
 \draw [red,thick,dashed] (3,3) to[out=30,in=100] (5.5,2);
\end{tikzpicture}
\end{document}

或者，如果你愿意，你可以算出抛物线到达下平面的时间，然后绘制出来。如果我没记错的话：

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
 \newcommand\initialV{4} % initial speed
 \newcommand\launchangle{30} % angle of launch
 \newcommand\gravity{9.81} % force of gravity

 % the lower plane is one unit below the exit, so set y = t*v_0*sin(theta) - 0.5*g*t^2 = -1, and solve for t
 \pgfmathsetmacro\timeofimpact{\initialV*sin(\launchangle)/\gravity + sqrt((\initialV*sin(\launchangle)/\gravity)^2 + 2/\gravity)}
 \draw  (0,3) -- (3,3)-- node[left] {H} (3,2) -- node[below] {$L_{\mathrm{max}}$} (6,2);
 \draw [dashed]   (3,3) -- +(0:1.5);
 \draw [->,>=stealth,line width=0.3mm]   (3,3) -- +(\launchangle:1.5) node [near end, above] {${\vec{v}}_0$};
 \draw (4,3) arc (0:30:1) node at (38:5.3) {$\alpha$};
 % domain based on calculation above
 \draw [red,thick,dashed,variable=\t,domain=0:\timeofimpact] (3,3) plot ({3+\t*\initialV*cos(\launchangle)},{3+\t*\initialV*sin(\launchangle) - 0.5*\gravity*\t^2});
\end{tikzpicture}
\end{document}