# 考虑空气阻力的竖直上抛问题

### 上升段

v'=-g-kv^2
\frac{dv}{dt}=-g-kv^2
\frac{dv}{g+kv^2}=-dt
\int{\frac{1}{g+kv^2}}dv=-t
w=\sqrt{\frac{k}{g}}v
\int{\frac{1}{g+kv^2}}dv
=\frac{1}{g}\int{\frac{1}{1+\frac{k}{g}v^2}}dv
=\frac{1}{g}\int{\frac{1}{1+wv^2}}dv
=\frac{1}{g}\int{\frac{1}{1+wv^2}}d\sqrt{\frac{g}{k}}w

=\frac{1}{g}\sqrt{\frac{g}{k}}tan^{-1}w+C
=\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v)+C

\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v)+C=-t
tan^{-1}(\sqrt{\frac{k}{g}}v)=-\sqrt{kg}(t+C)
\sqrt{\frac{k}{g}}v=tan(-\sqrt{kg}(t+C))
v=\sqrt{\frac{g}{k}}tan(-\sqrt{kg}(t+C))
C 带入
v=\sqrt{\frac{g}{k}}tan(-t\sqrt{kg}+tan^{-1}(\sqrt{\frac{k}{g}}v_0))

## 那么我们分个段吧

v=\sqrt{\frac{g}{k}}tan(-t_1\sqrt{kg}+tan^{-1}(\sqrt{\frac{k}{g}}v_0))=0
tan(-t_1\sqrt{kg}+tan^{-1}(\sqrt{\frac{k}{g}}v_0))=0

-t_1\sqrt{kg}+tan^{-1}(\sqrt{\frac{k}{g}}v_0)=0
tan^{-1}(\sqrt{\frac{k}{g}}v_0)=t_1\sqrt{kg}
t_1=\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v_0)

## 下降段

v'=-g+kv^2
\frac{dv}{dt}=-g+kv^2
\int{\frac{1}{g-kv^2}}dv=-t

=\frac{1}{\sqrt{gk}}tanh^{-1}(\sqrt{\frac{k}{g}}v)+C

\frac{1}{\sqrt{gk}}tanh^{-1}(\sqrt{\frac{k}{g}}v)+C=-t_1=-\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v_0)
v=0

C=-\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v_0)

v=\sqrt{\frac{g}{k}}tan(-t\sqrt{kg}+tanh^{-1}(\sqrt{\frac{k}{g}}v_0))

## OK 终极总结

v = \begin{cases} \sqrt{\frac{g}{k}}tan(-t\sqrt{kg}+tan^{-1}(\sqrt{\frac{k}{g}}v_0))&0\le t<\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v_0)\\\sqrt{\frac{g}{k}}tan(-t\sqrt{kg}+tanh^{-1}(\sqrt{\frac{k}{g}}v_0))&t\ge\frac{1}{\sqrt{gk}}tan^{-1}(\sqrt{\frac{k}{g}}v_0) \end{cases}

## Bonus!!!

Mathematica绘制的函数图象
k=0.5,g=10,v_0=10

Figure 1: v-t graph

Figure 2: s-t graph

THE END