题目描述
设一物体在无限高的地方以初速度为 $0$ 掉落，仅受到重力 $G=mg$，以及 zrq理想空气 阻力 $F=kv^2$
有若干次询问，每次给定 $t$，求在 $t$ 时刻该物体的速度和位移
最后一行输出该物体可能达到的最大速度
题解
先证明一个引理：
$$
\begin{aligned}
& \int \frac{dx}{a-bx^2} \\
=& \frac{1}{2\sqrt a}\int \left(\frac{dx}{\sqrt a-\sqrt bx} + \frac{dx}{\sqrt a+\sqrt bx}\right) \\
=& \frac{1}{2\sqrt a} \left(\frac{1}{-\sqrt b}\ln |\sqrt a-\sqrt bx|+\frac{1}{\sqrt b} \ln |\sqrt a + \sqrt bx| \right)
\end{aligned}
$$
然后有：
$$
\begin{aligned}
& ma=mg-kv^2 \\
\Rightarrow
& \frac{mdv}{dt}=mg-kv^2 \\
\Rightarrow
& \frac{mdv}{mg-kv^2}=dt \\
\Rightarrow
& m\int \frac{dv}{mg-kv^2} =C+t \\
\Rightarrow
& m \frac{\ln(\sqrt{mg}+\sqrt{k}v) - \ln(\sqrt{mg}-\sqrt{k}v)}{2\sqrt{mgk}}=C+t \\
\Rightarrow
& \ln(\sqrt{mg}+\sqrt{k}v) - \ln(\sqrt{mg}-\sqrt{k}v)=\frac{2\sqrt{mgk}(C+t)}{m} \\
\Rightarrow
& \frac{\sqrt{mg}+\sqrt{k}v}{\sqrt{mg}-\sqrt{k}v}=e^{\frac{2\sqrt{mgk}(C+t)}{m}} \\
\Rightarrow
& \sqrt{mg}+\sqrt{k}v=e^{\frac{2\sqrt{mgk}(C+t)}{m}}(\sqrt{mg}-\sqrt{k}v) \\
\Rightarrow
& v=\frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}(C+t)}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}(C+t)}{m}}+1\right)}
\end{aligned}
$$
由于 $t=0$ 时，有 $v=0$，于是：
$$
\begin{aligned}
& 0=\frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}C}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}C}{m}}+1\right)} \\
\Rightarrow
& 1=e^{\frac{2\sqrt{mgk}C}{m}} \\
\Rightarrow
& 0=\frac{2\sqrt{mgk}C}{m} \\
\Rightarrow
& C=0
\end{aligned}
$$
因此：
$$
v=\frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}t}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}t}{m}}+1\right)}
$$
同时有：
$$
\begin{aligned}
v_{\max}
=& \lim_{t \to \infty}\frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}t}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}t}{m}}+1\right)} \\
=& \sqrt \frac{mg}{k}\lim_{t \to \infty}\frac{2\sqrt{mgk}e^{\frac{2\sqrt{mgk}t}{m}}}{2\sqrt{mgk}e^{\frac{2\sqrt{mgk}t}{m}}} \\
=&\sqrt \frac{mg}{k}
\end{aligned}
$$
还要证明另一个引理：
$$
\int \frac{e^x-1}{e^{x}+1}dx=2 \ln(e^x+1)-x
$$
因为有：
$$
v=\frac{dx}{dt}
$$
于是：
$$
\begin{aligned}
& \frac{dx}{dt} = \frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}t}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}t}{m}}+1\right)} \\
\Rightarrow
& x=C+\int \frac{\sqrt{mg}\left(e^{\frac{2\sqrt{mgk}t}{m}}-1\right)}{\sqrt k\left(e^{\frac{2\sqrt{mgk}t}{m}}+1\right)} dt \\
\Rightarrow
& x=C+\sqrt \frac{mg}{k} \frac{2 \ln (e^{2\sqrt{mgk}t}+1)-2\sqrt{mgk}t}{2\sqrt{mgk}} \\
\Rightarrow
& x= C+\frac{\ln(e^{2\sqrt{mgk}t}+1)-\sqrt{mgk}t}{k}\\
\end{aligned}
$$
显然 $C=-\frac{\ln 2}{k}$，所以：
$$
x= \frac{\ln(e^{2\sqrt{mgk}t}+1)-\sqrt{mgk}t}{k}-\frac{\ln 2}{k}
$$
