带空气阻力的抛射体飞行运动轨迹
带空气阻力的抛射体飞行运动轨迹
Write By Champrin From 2022-11-20 To 2022-11-
GUET Evolution Team Visual Group
目录
在低速情况下，常用的空气阻力模型有两种：
- 空气阻力与速度的一次方成正比
- 空气阻力与速度的二次方成正比
论文《两种空气阻力模型的抛射体飞行轨迹研究 闵永林 陈池》指出：
- 物体的速度低于\(200m/s\)时，可认为阻力与物体速度大小的平方成正比
- 物体速度远小于低速炮弹(\(300m/s\))时，采用第一种模型；物体速度接近于低速炮弹时,采用第二种模型
对于42mm弹丸和飞镖的初射速都低于这两个值，因此对于文章中指出的，两种模型是否都能采用，需要进行验证采用哪种模型效果更好
在上赛季英雄视觉代码中，建立的是对42mm弹丸忽略了竖直方向上的空气阻力的方程，这或许对短距离来说影响不大，但是若要英雄进行远距离吊射，以及用于飞镖视觉当中，需要更高的准确性，需要建立更精确和更细致的方程
空气阻力与速度的一次方成正比
由牛顿第二定律：
\[\begin{aligned}
ma_x & = - kv\cos{\theta} \\
ma_y & = - (mg + kv\sin{\theta})
\end{aligned}
\]
也即：
\[\begin{aligned}
m\frac{d^2x}{d^2t} & = - k\frac{dx}{dt} \\
m\frac{d^2y}{d^2t} & = -mg - k\frac{dy}{dt}
\end{aligned}
\]
也即：
\[\begin{aligned}
m\frac{dv_x}{dt} & = - kv_x & \text{(1)}\\
m\frac{dv_y}{dt} & = -mg - kv_y & \text{(2)}
\end{aligned}
\]
速度与位移方程
对\(\text{(1)}\)式分离变量，且两边同时积分，同时\(v_x > 0\)：
\[\begin{aligned}
\frac{dv_x}{v_x} & = - \frac{k}{m}dt \\
\int \frac{dv_x}{v_x} & = \int - \frac{k}{m}dt \\
\Rightarrow \ln{v_x} + C_{v_x} & = - \frac{k}{m}t
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_{v_x}\)：
\[ C_{v_x} = -\ln{v_x} = -\ln{v_0\cos{\theta}}
\]
可得\(v_x\)：
\[\begin{aligned}
\ln{v_x} -\ln{v_0\cos{\theta}} & = - \frac{k}{m}t \\
\ln\frac{v_x}{v_0\cos{\theta}} & = - \frac{k}{m}t \\
\frac{v_x}{v_0\cos{\theta}} & = e^{- \frac{k}{m}t} \\
\Rightarrow v_x & = v_0\cos{\theta} \cdot e^{- \frac{k}{m}t} \qquad \text{(3)}
\end{aligned}
\]
对于\(v_x\)，有：
\[\begin{aligned}
v_x & = \frac{dx}{dt} \\
v_0\cos{\theta} \cdot e^{- \frac{k}{m}t} & = \frac{dx}{dt}
\end{aligned}
\]
分离变量，两边同时积分得：
\[\begin{aligned}
\int dx & = \int v_0\cos{\theta} \cdot e^{- \frac{k}{m}t} dt \\
\int dx & = \int - \frac{m}{k} \cdot v_0\cos{\theta} \cdot e^{- \frac{k}{m}t} d(- \frac{k}{m}t) \\
\Rightarrow x + C_x & = - \frac{mv_0\cos{\theta}}{k}e^{- \frac{k}{m}t}
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_x\)：
\[\begin{aligned}
x_0 + C_x & = - \frac{mv_0\cos{\theta}}{k} \\
C_x & = - \frac{mv_0\cos{\theta}}{k} - x_0
\end{aligned}
\]
可得\(x\)：
\[\begin{aligned}
x - \frac{mv_0\cos{\theta}}{k} - x_0 & = - \frac{mv_0\cos{\theta}}{k}e^{- \frac{k}{m}t} \\
x & = x_0 - \frac{mv_0\cos{\theta}}{k}e^{- \frac{k}{m}t} + \frac{mv_0\cos{\theta}}{k} \\
\Rightarrow x & = x_0 + \frac{mv_0\cos{\theta}}{k}(1 - e^{- \frac{k}{m}t}) \qquad \text{(4)}
\end{aligned}
\]
同理，对\(\text{(2)}\)式分离变量，两边同时积分：
\[\begin{aligned}
\frac{dv_y}{-g - \frac{k}{m}v_y} & = dt \\
\int \frac{dv_y}{-g - \frac{k}{m}v_y} & = \int dt \\
\int -\frac{m}{k}\frac{d(-g - \frac{k}{m}v_y)}{-g - \frac{k}{m}v_y} & = \int dt \\
\Rightarrow -\frac{m}{k}[\ln{(-g - \frac{k}{m}v_y)} + C_{v_y}] & = t
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_{v_y}\)：
\[ C_{v_y} = -\ln{(-g - \frac{k}{m}v_y)} = -\ln{(-g - \frac{k}{m}v_0\sin{\theta})}
\]
可得\(v_y\)：
\[\begin{aligned}
-\frac{m}{k}[\ln{(-g - \frac{k}{m}v_y)} -\ln{(-g - \frac{k}{m}v_0\sin{\theta})}] & = t \\
\ln(\frac{-g - \frac{k}{m}v_y}{-g - \frac{k}{m}v_0\sin{\theta}}) & = -\frac{k}{m}t \\
\frac{-g - \frac{k}{m}v_y}{-g - \frac{k}{m}v_0\sin{\theta}} & = e^{-\frac{k}{m}t}\\
-g - \frac{k}{m}v_y & = (-g - \frac{k}{m}v_0\sin{\theta})e^{-\frac{k}{m}t} \\
v_y & = \frac{(-g - \frac{k}{m}v_0\sin{\theta})e^{-\frac{k}{m}t} + g}{- \frac{k}{m}} \\
\Rightarrow v_y & = (\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k} \qquad \text{(5)}
\end{aligned}
\]
对于\(v_y\)，有：
\[\begin{aligned}
v_y & = \frac{dy}{dt} \\
(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k} & = \frac{dy}{dt}
\end{aligned}
\]
分离变量，两边同时积分得：
\[\begin{aligned}
\int dy & = \int [(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k}]dt \\
\int dy & = \int (\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t}dt - \int \frac{mg}{k}dt \\
\int dy & = \int - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t}d(-\frac{k}{m}t) - \int \frac{mg}{k}dt \\
\Rightarrow y + C_y & = - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k}t
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_y\)：
\[\begin{aligned}
+ y_0C_y = - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta}) \\
C_y = - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta}) - y_0
\end{aligned}
\]
可得\(y\)：
\[\begin{aligned}
y - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta}) - y_0 & = - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k}t \\
y & = y_0 - \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k}t + \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta}) \\
y & = y_0 + \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(1 -e^{-\frac{k}{m}t}) - \frac{mg}{k}t \qquad \text{(6)}
\end{aligned}
\]
轨迹方程
由\(\text{(4)}\)式可得：
\[\begin{aligned}
x & = x_0 + \frac{mv_0\cos{\theta}}{k}(1 - e^{- \frac{k}{m}t}) \\
1 - e^{- \frac{k}{m}t} & = \frac{k(x - x_0)}{mv_0\cos{\theta}} \qquad & \text{(7)} \\
e^{- \frac{k}{m}t} & = 1 - \frac{k(x - x_0)}{mv_0\cos{\theta}} \\
t & = \ln(1 - \frac{k(x - x_0)}{mv_0\cos{\theta}}) \\
t & = - \frac{m}{k}\ln(1 - \frac{k(x - x_0)}{mv_0\cos{\theta}}) \qquad & \text{(8)}
\end{aligned}
\]
联立\(\text{(6)}\text{(7)}\text{(8)}\)式，消去\(t\)，可得轨迹方程：
\[\begin{aligned}
y & = y_0 + \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(1 -e^{-\frac{k}{m}t}) - \frac{mg}{k}t \\
y & = y_0 + \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})\frac{k(x - x_0)}{mv_0\cos{\theta}} + \frac{mg}{k}\frac{m}{k}\ln(1 - \frac{k(x - x_0)}{mv_0\cos{\theta}}) \\
y & = y_0 + (\frac{mg}{k} + v_0\sin{\theta})\frac{(x - x_0)}{v_0\cos{\theta}} + \frac{mg}{k}\frac{m}{k}\ln(1 - \frac{k(x - x_0)}{mv_0\cos{\theta}}) \\
y & = y_0 + (\frac{mg}{kv_0\cos{\theta}} + \tan{\theta})(x - x_0) + \frac{m^2g}{k^2}\ln(1 - \frac{k(x - x_0)}{mv_0\cos{\theta}}) \qquad \text{(9)}
\end{aligned}
\]
令\(v_y = 0\)，由式\((3)\)可得抛射体到达最高点的时间：
\[\begin{aligned}
0 & = (\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} - \frac{mg}{k} \\
(\frac{mg}{k} + v_0\sin{\theta})e^{-\frac{k}{m}t} & = \frac{mg}{k} \\
e^{-\frac{k}{m}t} & = \cfrac{\cfrac{mg}{k}}{\cfrac{mg}{k} + v_0\sin{\theta}} \\
-\frac{k}{m}t & = \ln{(\cfrac{\cfrac{mg}{k}}{\cfrac{mg}{k} + v_0\sin{\theta}})} \\
t & = -\frac{m}{k}\ln{(\cfrac{mg}{mg + kv_0\sin{\theta}})} \\
\end{aligned}
\]
落回初始高度所需时间\(t\)
\(y = y_0\)为初始高度，那么令\(\text{(6)}\)式中\(y = y_0\)得：
\[\begin{aligned}
y_0 & = y_0 + \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(1 -e^{-\frac{k}{m}t}) - \frac{mg}{k}t \\
0 & = \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(1 -e^{-\frac{k}{m}t}) - \frac{mg}{k}t \\
0 & = (\frac{mg}{k} + v_0\sin{\theta})(1 - e^{-\frac{k}{m}t}) - gt \\
gt & = (\frac{mg}{k} + v_0\sin{\theta})(1 - e^{-\frac{k}{m}t})\qquad \text{(10)}
\end{aligned}
\]
可见，这是一个超越方程，无法直接求出\(t\)，那么可以想到展开\(e^{-\frac{k}{m}t}\)
展开\(e^{-\frac{k}{m}t}\)
在\(t = 0\)处，把\(e^{-\frac{k}{m}t}\)展开成\(t\)的幂级数：
第一步：\[f(0) = 1 ,f'(0) = (-\frac{k}{m})^1 ,f''(0) = (-\frac{k}{m})> ^2, ... ,f^{(n)}(0) = (-\frac{k}{m})^n \]第二步：
\[ f(0) + f'(0)t + \frac{f''(0)}{2!}t^2 + ... + \frac{f^{(n)} (0)}{n!}t^n + ... \]得级数：
\[ 1 + (-\frac{k}{m})t + \frac{(-\frac{k}{m})^2}{2!}t^2 > + ... + \frac{(-\frac{k}{m})^n}{n!}t^n + ... \]\[ \sum_{n=0}^{\infty}\frac{(-\frac{k}{m})^n}{n!}t^n \]求此级数收敛半径：
\[ \rho = \lim_{n \rarr \infty} \lvert \frac{a_{n + 1}}{a_n} \rvert = \lim_{n \rarr \infty} \lvert \frac{\frac{(-\frac{k}{m})^{n+1}}{(n+1)!}}{\frac{(-\frac{k}{m})^n}{n!}} \rvert = \lim_{n \rarr \infty} \lvert -\frac{k}{m(n + 1)} \rvert = 0 \]因此，收敛半径\(R\)：
\[ R = + \infty \]进而利用余项\(R_n(t)\)的表达式，求得当\(t \in (-R, R)\)内时，余项\(R_n(t)\)的极限为0
因此：\[ e^{-\frac{k}{m}t} = \sum_{n=0}^{\infty}\frac{(-\frac{k}{m})^n}{n!}t^n \qquad (-\infty \lt t \lt +\infty) \]
取展开式的二次项：
\[ e^{-\frac{k}{m}t} \approx 1 - \frac{k}{m}t + \frac{1}{2}(\frac{k}{m})^2t^2 \qquad \text{(11)}
\]
联立\(\text{(10)}\text{(11)}\)式，得：
\[\begin{aligned}
gt & = (\frac{mg}{k} + v_0\sin{\theta})(1 - 1 + \frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2) \\
gt & = (\frac{mg}{k} + v_0\sin{\theta})(\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2) \\
g & = (\frac{mg}{k} + v_0\sin{\theta})(\frac{k}{m} - \frac{1}{2}(\frac{k}{m})^2t) \\
g & = (\frac{mg}{k} + v_0\sin{\theta})\frac{k}{m} - (\frac{mg}{k} + v_0\sin{\theta})\frac{1}{2}(\frac{k}{m})^2t \\
(\frac{mg}{k} + v_0\sin{\theta})\frac{1}{2}(\frac{k}{m})^2t & = (\frac{mg}{k} + v_0\sin{\theta})\frac{k}{m} - g \\
t & = \frac{(\frac{mg}{k} + v_0\sin{\theta})\frac{k}{m} - g}{(\frac{mg}{k} + v_0\sin{\theta})\frac{1}{2}(\frac{k}{m})^2} \\
t & = \frac{g + \frac{kv_0\sin{\theta}}{m} - g}{(\frac{mg}{k} + v_0\sin{\theta})\frac{1}{2}(\frac{k}{m})^2} \\
t & = \frac{\frac{kv_0\sin{\theta}}{m}}{(\frac{mg}{k} + v_0\sin{\theta})\frac{1}{2}(\frac{k}{m})^2} \\
t & = \frac{2v_0\sin{\theta}}{(\frac{mg}{k} + v_0\sin{\theta})\frac{k}{m}} \\
t & = \frac{2mv_0\sin{\theta}}{mg + kv_0\sin{\theta}} \qquad \text{(12)}\\
\end{aligned}
\]
最大射程及其对应\(\theta\)
将\(\text{(11)}\text{(12)}\)式带入\(\text{(4)}\)，得：
\[\begin{aligned}
x & = x_0 + \frac{mv_0\cos{\theta}}{k}(1 - e^{- \frac{k}{m}t}) \\
& = x_0 + \frac{mv_0\cos{\theta}}{k}(1 - 1 + \frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2) \\
& = x_0 + v_0\cos{\theta}(t - \frac{1}{2}\frac{k}{m}t^2) \\
& = x_0 + v_0\cos{\theta} \cdot t \cdot (1 - \frac{k}{2m}t) \\
& = x_0 + v_0\cos{\theta} \frac{2mv_0\sin{\theta}}{mg + kv_0\sin{\theta}} (1 - \frac{k}{2m}\frac{2mv_0\sin{\theta}}{mg + kv_0\sin{\theta}}) \\
& = x_0 + \frac{2mv^2_0\sin{\theta}\cos{\theta}}{mg + kv_0\sin{\theta}} (1 - \frac{kv_0\sin{\theta}}{mg + kv_0\sin{\theta}}) \\
& = x_0 + \frac{2mv^2_0\sin{\theta}\cos{\theta}}{mg + kv_0\sin{\theta}} (\frac{mg + kv_0\sin{\theta}}{mg + kv_0\sin{\theta}} - \frac{kv_0\sin{\theta}}{mg + kv_0\sin{\theta}}) \\
& = x_0 + \frac{2mv^2_0\sin{\theta}\cos{\theta}}{mg + kv_0\sin{\theta}} \frac{mg}{mg + kv_0\sin{\theta}} \\
& = x_0 + \frac{m^2gv^2_0 \cdot 2\sin{\theta}\cos{\theta}}{(mg + kv_0\sin{\theta})^2} \\
& = x_0 + \frac{m^2gv^2_0\sin{2\theta}}{(mg + kv_0\sin{\theta})^2} \qquad \text{(13)}\\
\end{aligned}
\]
要求最大射程，即求\(x\)的最大值，对\(\text{(13)}\)式利用求导法：\(\frac{dx}{d\theta} = 0\)，求得\(x\)的最大值，及其对应\(\theta\)
\[\begin{aligned}
\frac{dx}{d\theta} & = m^2gv^2_0(\frac{\sin{2\theta}}{(mg + kv_0\sin{\theta})^2})' \\
& = m^2gv^2_0(\frac{(\sin{2\theta})'(mg + kv_0\sin{\theta})^2 - \sin{2\theta}((mg + kv_0\sin{\theta})^2)'}{(mg + kv_0\sin{\theta})^4}) \\
& = m^2gv^2_0(\frac{2\cos{2\theta}(mg + kv_0\sin{\theta})^2 - \sin{2\theta}(2kv_0\cos{\theta}(mg + kv_0\sin{\theta}))}{(mg + kv_0\sin{\theta})^4}) \\
& = m^2gv^2_0(\frac{2\cos{2\theta}(mg + kv_0\sin{\theta})^2 - 2kv_0\sin{2\theta}\cos{\theta}(mg + kv_0\sin{\theta})}{(mg + kv_0\sin{\theta})^4})
\end{aligned}
\]
\[ 2\cos{2\theta}(mg + kv_0\sin{\theta})^2 - 2kv_0\sin{2\theta}\cos{\theta}(mg + kv_0\sin{\theta}) = 0 \\
\cos{2\theta}(mg + kv_0\sin{\theta})^2 - kv_0\sin{2\theta}\cos{\theta}(mg + kv_0\sin{\theta}) = 0 \\
\Rightarrow 2mgv^2_0\cos^2\theta + b\cos\theta - mgv^2_0 = 0
\]
空气阻力与速度的二次方成正比
由牛顿第二定律：
\[\begin{aligned}
ma_x & = - kv^2_x \\
ma_y & = - (mg + kv^2_y)
\end{aligned}
\]
也即：
\[\begin{aligned}
m\frac{dv_x}{dt} & = - kv^2_x & \text{(1)}\\
m\frac{dv_y}{dt} & = -mg - kv^2_y & \text{(2)} \\
\end{aligned} \\
\]
速度与位移方程
对\(\text{(1)}\)式分离变量，两边同时积分：
\[\begin{aligned}
\frac{dv_x}{v^2_x} & = - \frac{k}{m}dt \\
\int \frac{dv_x}{v^2_x} & = \int - \frac{k}{m}dt \\
- \frac{1}{v_x} + C_{v_x} & = - \frac{k}{m}t
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_{v_x}\)：
\[ C_{v_x} = \frac{1}{v_x} = \frac{1}{v_0\cos{\theta}}
\]
可得\(v_x\)：
\[\begin{aligned}
- \frac{1}{v_x} + \frac{1}{v_0\cos{\theta}} & = - \frac{k}{m}t \\
\frac{1}{v_x} & = \frac{1}{v_0\cos{\theta}} + \frac{k}{m}t \\
v_x & = \frac{1}{\frac{1}{v_0\cos{\theta}} + \frac{k}{m}t} \\
\Rightarrow v_x & = \frac{mv_0\cos{\theta}}{m + kv_0\cos{\theta}t} \qquad \text{(3)}
\end{aligned}
\]
对于\(v_x\)，有：
\[\begin{aligned}
v_x & = \frac{dx}{dt} \\
\frac{mv_0\cos{\theta}}{m + kv_0\cos{\theta}t} & = \frac{dx}{dt}
\end{aligned}
\]
分离变量，两边同时积分得：
\[\begin{aligned}
\frac{dx}{mv_0\cos{\theta}} & = \frac{dt}{m + kv_0\cos{\theta}t} \\
\int \frac{dx}{mv_0\cos{\theta}} & = \int \frac{dt}{m + kv_0\cos{\theta}t} \\
\int \frac{dx}{mv_0\cos{\theta}} & = \int \frac{1}{kv_0\cos{\theta}}\frac{d(m + kv_0\cos{\theta}t)}{m + kv_0\cos{\theta}t} \\
\Rightarrow \frac{1}{mv_0\cos{\theta}}x + C_x & = \frac{1}{kv_0\cos{\theta}} \ln(m + kv_0\cos{\theta}t)
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_x\)：
\[\begin{aligned}
\frac{1}{mv_0\cos{\theta}}x_0 + C_x & = \frac{1}{kv_0\cos{\theta}}\ln{m} \\
C_x & = \frac{1}{kv_0\cos{\theta}}\ln{m} - \frac{1}{mv_0\cos{\theta}}x_0\\
\end{aligned}
\]
可得\(x\)：
\[\begin{aligned}
\frac{1}{mv_0\cos{\theta}}x + \frac{1}{kv_0\cos{\theta}}\ln{m} - \frac{1}{mv_0\cos{\theta}}x_0= \frac{1}{kv_0\cos{\theta}} \ln(m + kv_0\cos{\theta}t) \\
\frac{1}{mv_0\cos{\theta}}x = \frac{1}{kv_0\cos{\theta}} \ln(m + kv_0\cos{\theta}t) - \frac{1}{kv_0\cos{\theta}}\ln{m} + \frac{1}{mv_0\cos{\theta}}x_0 \\
\frac{1}{mv_0\cos{\theta}}x = \frac{1}{kv_0\cos{\theta}}(\ln(m + kv_0\cos{\theta}t) - \ln{m}) + \frac{1}{mv_0\cos{\theta}}x_0\\
\frac{1}{mv_0\cos{\theta}}x = \frac{1}{kv_0\cos{\theta}}\ln\frac{(m + kv_0\cos{\theta}t)}{m} + \frac{1}{mv_0\cos{\theta}}x_0\\
\Rightarrow x = \frac{m}{k}\ln(\frac{m + kv_0\cos{\theta}t}{m}) + x_0\qquad \text{(4)}
\end{aligned}
\]
同理，对\(\text{(2)}\)式分离变量和同时积分得，
\[\begin{aligned}
m\frac{dv_y}{dt} = -mg - kv^2_y \\
\frac{dv_y}{dt} = -g - \frac{k}{m}v^2_y \\
\frac{dv_y}{-g - \frac{k}{m}v^2_y} = dt
\end{aligned}
\]
两边同时积分：
\[\begin{aligned}
\int \frac{dv_y}{-g - \frac{k}{m}v^2_y} & = \int dt \\
\int \frac{dv_y}{-g - \frac{k}{m}v^2_y} & = \int dt \\
\int -\frac{mdv_y}{mg + kv^2_y} & = \int dt \\
-m\int \frac{dv_y}{mg + kv^2_y} & = \int dt \\
-m\int \frac{dv_y}{(\sqrt{mg})^2 + (\sqrt{k}v_y)^2} & = \int dt \\
-m\int \frac{dv_y}{1 + (\frac{\sqrt{k}v_y}{\sqrt{mg}})^2} & = \int dt \\
-m\frac{1}{(\sqrt{mg})^2}\int \frac{\sqrt{mg}}{\sqrt{k}}\frac{d(\frac{\sqrt{k}v_y}{\sqrt{mg}})}{1 + (\frac{\sqrt{k}v_y}{\sqrt{mg}})^2} & = \int dt \\
-m\frac{1}{(\sqrt{mg})^2}\frac{\sqrt{mg}}{\sqrt{k}}\int \frac{d(\frac{\sqrt{k}v_y}{\sqrt{mg}})}{1 + (\frac{\sqrt{k}v_y}{\sqrt{mg}})^2} & = \int dt \\
-m\frac{1}{(\sqrt{mg})^2}\frac{\sqrt{mg}}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = \int dt \\
-m\frac{1}{\sqrt{mg}}\frac{1}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = \int dt \\
-m\frac{\sqrt{mg}}{mg}\frac{1}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = \int dt \\
-\frac{\sqrt{mg}}{g\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = \int dt \\
\end{aligned}
\]
得：
\[\begin{aligned}
-m\frac{1}{(\sqrt{mg})^2}\frac{\sqrt{mg}}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = t \\
-m\frac{1}{\sqrt{mg}}\frac{1}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = t \\
-m\frac{\sqrt{mg}}{mg}\frac{1}{\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = t \\
-\frac{\sqrt{mg}}{g\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) + C_{v_y}) & = t \\
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_{v_y}\)：
\[ C_{v_y} = - \arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) = - \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})
\]
注意到：
\[ \arctan A - \arctan B = \arctan\frac{A-B}{1 + AB} \\
\tan (\arctan A) = A
\]
可得\(v_y\)：
\[\begin{aligned}
-\frac{\sqrt{mg}}{g\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) - \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})) & = t \\
\arctan\frac{(\frac{\sqrt{k}v_y}{\sqrt{mg}}) - (\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})}{1 + (\frac{\sqrt{k}v_y}{\sqrt{mg}})(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})} & = -\frac{tg\sqrt{k}}{\sqrt{mg}} \\
\arctan\frac{\frac{\sqrt{k}}{\sqrt{mg}}(v_y - v_0\sin{\theta})}{1 + \frac{kv_yv_0\sin{\theta}}{mg}} & = -\frac{tg\sqrt{k}}{\sqrt{mg}} \\
\tan(\arctan\frac{\frac{\sqrt{k}}{\sqrt{mg}}(v_y - v_0\sin{\theta})}{1 + \frac{kv_yv_0\sin{\theta}}{mg}}) & = \tan(-\frac{tg\sqrt{k}}{\sqrt{mg}}) \\
\frac{\frac{\sqrt{k}}{\sqrt{mg}}(v_y - v_0\sin{\theta})}{1 + \frac{kv_yv_0\sin{\theta}}{mg}} & = \tan(-\frac{tg\sqrt{k}}{\sqrt{mg}}) \\
\frac{mg\frac{\sqrt{k}}{\sqrt{mg}}(v_y - v_0\sin{\theta})}{mg + kv_yv_0\sin{\theta}} & = \tan(-\frac{tg\sqrt{k}}{\sqrt{mg}}) \\
\frac{\sqrt{mgk}(v_y - v_0\sin{\theta})}{mg + kv_yv_0\sin{\theta}} & = \tan(-\frac{t\sqrt{mgk}}{m}) \\
\sqrt{mgk}(v_y - v_0\sin{\theta}) & = \tan(-\frac{t\sqrt{mgk}}{m})(mg + kv_yv_0\sin{\theta}) \\
\sqrt{mgk}v_y - \sqrt{mgk}v_0\sin{\theta} & = \tan(-\frac{t\sqrt{mgk}}{m})mg + \tan(-\frac{t\sqrt{mgk}}{m})kv_yv_0\sin{\theta} \\
\sqrt{mgk}v_y - \tan(-\frac{t\sqrt{mgk}}{m})kv_yv_0\sin{\theta} & = \tan(-\frac{t\sqrt{mgk}}{m})mg + \sqrt{mgk}v_0\sin{\theta} \\
v_y(\sqrt{mgk} - \tan(-\frac{t\sqrt{mgk}}{m})kv_0\sin{\theta}) & = \tan(-\frac{t\sqrt{mgk}}{m})mg + \sqrt{mgk}v_0\sin{\theta} \\
\Rightarrow v_y & = \frac{mg\tan(-\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} - kv_0\sin{\theta}\tan(-\frac{t\sqrt{mgk}}{m})} \qquad \text{(5.1)}
\end{aligned}
\]
同时，还有另外一种形式：
\[\begin{aligned}
-\frac{\sqrt{mg}}{g\sqrt{k}}(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) - \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})) & = t \\
\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) - \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}}) & = -\frac{tg\sqrt{k}}{\sqrt{mg}} \\
\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) - \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}}) & = -t\sqrt{\frac{gk}{m}} \\
\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}}) & = -t\sqrt{\frac{gk}{m}} + \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}})\\
\tan(\arctan(\frac{\sqrt{k}v_y}{\sqrt{mg}})) & = \tan(-t\sqrt{\frac{gk}{m}} + \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}}))\\
\frac{\sqrt{k}v_y}{\sqrt{mg}} & = \tan(-t\sqrt{\frac{gk}{m}} + \arctan(\frac{\sqrt{k}v_0\sin{\theta}}{\sqrt{mg}}))\\
\Rightarrow v_y & = \sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})) \qquad \text{(5.2)}
\end{aligned}
\]
对于\(v_y\text{(5.1)}\)，有：
\[\begin{aligned}
v_y & = \frac{dy}{dt} \\
\frac{mg\tan(-\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} - kv_0\sin{\theta}\tan(-\frac{t\sqrt{mgk}}{m})} & = \frac{dy}{dt}
\end{aligned}
\]
分离变量，两边同时积分得：
\[\begin{aligned}
dy & = \frac{mg\tan(-\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} - kv_0\sin{\theta}\tan(-\frac{t\sqrt{mgk}}{m})}dt \\
\int dy & = \int \frac{mg\tan(-\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} - kv_0\sin{\theta}\tan(-\frac{t\sqrt{mgk}}{m})}dt \\
\int dy & = \int \frac{-mg\tan(\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} + kv_0\sin{\theta}\tan(\frac{t\sqrt{mgk}}{m})}dt \\
\end{aligned}
\]
对右边单独处理：
\[\begin{aligned}
& 令u = \frac{t\sqrt{mgk}}{m}, 则t = \frac{mu}{\sqrt{mgk}}, dt = \frac{m}{\sqrt{mgk}}du \\
& \int \frac{-mg\tan u + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}\frac{m}{\sqrt{mgk}}du \\
& = \frac{m}{\sqrt{mgk}}\int \frac{-mg\tan u + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}du \\
提取分数的整数部分 \\
& = \frac{m}{\sqrt{mgk}}\int (\frac{kv^2_0\sin^2{\theta}\sqrt{mgk} + mg\sqrt{mgk}}{kv_0\sin{\theta}(\sqrt{mgk} + kv_0\sin{\theta}\tan u)} - \frac{mg}{kv_0\sin{\theta}})du \\
& = \frac{m}{\sqrt{mgk}}(\int \frac{kv^2_0\sin^2{\theta}\sqrt{mgk} + mg\sqrt{mgk}}{kv_0\sin{\theta}(\sqrt{mgk} + kv_0\sin{\theta}\tan u)}du - \int \frac{mg}{kv_0\sin{\theta}}du) \\
& = \frac{m}{\sqrt{mgk}}(\frac{kv^2_0\sin^2{\theta}\sqrt{mgk} + mg\sqrt{mgk}}{kv_0\sin{\theta}}\int \frac{1}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}du - \frac{mg}{kv_0\sin{\theta}}\int du) \\
\end{aligned}
\]
对\(\int \frac{1}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}du\)单独处理：
\[\begin{aligned}
& \int \frac{1}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}du \\
& = \int \frac{1}{\sqrt{mgk} + kv_0\sin{\theta}\frac{\sin u}{\cos u}}du \\
& = \int \frac{\cos u}{\cos u\sqrt{mgk} + kv_0\sin{\theta}\sin u}du \\
分组\cos u：\\
& \cos u = \frac{\sqrt{mgk}(kv_0\sin\theta\sin u + \sqrt{mgk}\cos u)}{mgk + k^2v^2_0\sin^2{\theta}} + \frac{kv_0\sin\theta(kv_0\sin\theta\cos u - \sqrt{mgk}\sin u)}{mgk + k^2v^2_0\sin^2{\theta}} \\
& = \int \frac{1}{\cos u\sqrt{mgk} + kv_0\sin{\theta}\sin u}(\frac{\sqrt{mgk}(kv_0\sin\theta\sin u + \sqrt{mgk}\cos u)}{mgk + k^2v^2_0\sin^2{\theta}} + \frac{kv_0\sin\theta(kv_0\sin\theta\cos u - \sqrt{mgk}\sin u)}{mgk + k^2v^2_0\sin^2{\theta}})du \\
& = \int (\frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}} + \frac{kv_0\sin\theta(kv_0\sin\theta\cos u - \sqrt{mgk}\sin u)}{(mgk + k^2v^2_0\sin^2{\theta})(\cos u\sqrt{mgk} + kv_0\sin{\theta}\sin u)})du \\
& = \int \frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}du + \int \frac{kv_0\sin\theta(kv_0\sin\theta\cos u - \sqrt{mgk}\sin u)}{(mgk + k^2v^2_0\sin^2{\theta})(\cos u\sqrt{mgk} + kv_0\sin{\theta}\sin u)}du \\
& = \frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}\int du + \frac{kv_0\sin\theta}{mgk + k^2v^2_0\sin^2{\theta}}\int \frac{kv_0\sin\theta\cos u - \sqrt{mgk}\sin u}{kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u}du \\
& 因为(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)' = kv_0\sin\theta\cos u - \sqrt{mgk}\sin u \\
& = \frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}\int du + \frac{kv_0\sin\theta}{mgk + k^2v^2_0\sin^2{\theta}}\int \frac{1}{kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u}d(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u) \\
& = \frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}u + \frac{v_0\sin\theta\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{mg + kv^2_0\sin^2{\theta}} \\
\end{aligned}
\]
因此：
\[\begin{aligned}
\int dy & = \frac{m}{\sqrt{mgk}}(\frac{kv^2_0\sin^2{\theta}\sqrt{mgk} + mg\sqrt{mgk}}{kv_0\sin{\theta}}\int \frac{1}{\sqrt{mgk} + kv_0\sin{\theta}\tan u}du - \frac{mg}{kv_0\sin{\theta}}\int du)\\
y & = \frac{m}{\sqrt{mgk}}[\frac{kv^2_0\sin^2{\theta}\sqrt{mgk} + mg\sqrt{mgk}}{kv_0\sin{\theta}}(\frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}u + \frac{v_0\sin\theta\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{mg + kv^2_0\sin^2{\theta}}) - \frac{mg}{kv_0\sin{\theta}}u + C_y] \\
y & = \frac{m}{\sqrt{mgk}}[\frac{\sqrt{mgk}(kv^2_0\sin^2{\theta} + mg)}{kv_0\sin{\theta}}(\frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}u + \frac{v_0\sin\theta\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{mg + kv^2_0\sin^2{\theta}}) - \frac{mg}{kv_0\sin{\theta}}u + C_y] \\
y & = \frac{m}{\sqrt{mgk}}(\frac{\sqrt{mgk}(kv^2_0\sin^2{\theta} + mg)}{kv_0\sin{\theta}}\frac{\sqrt{mgk}}{mgk + k^2v^2_0\sin^2{\theta}}u + \frac{\sqrt{mgk}(kv^2_0\sin^2{\theta} + mg)}{kv_0\sin{\theta}}\frac{v_0\sin\theta\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{mg + kv^2_0\sin^2{\theta}} - \frac{mg}{kv_0\sin{\theta}}u + C_y) \\
y & = \frac{m}{\sqrt{mgk}}(\frac{mgk(kv^2_0\sin^2{\theta} + mg)}{kv_0\sin{\theta}(mgk + k^2v^2_0\sin^2{\theta})}u + \frac{\sqrt{mgk}\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} - \frac{mg}{kv_0\sin{\theta}}u + C_y) \\
y & = \frac{m}{\sqrt{mgk}}(\frac{mg(k^2v^2_0\sin^2{\theta} + mgk)}{kv_0\sin{\theta}(mgk + k^2v^2_0\sin^2{\theta})}u + \frac{\sqrt{mgk}\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} - \frac{mg}{kv_0\sin{\theta}}u + C_y) \\
y & = \frac{m}{\sqrt{mgk}}(\frac{mg}{kv_0\sin{\theta}}u + \frac{\sqrt{mgk}\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} - \frac{mg}{kv_0\sin{\theta}}u + C_y) \\
y & = \frac{m}{\sqrt{mgk}}(\frac{\sqrt{mgk}\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} + C_y) \\
y & = \frac{m}{\sqrt{mgk}}\frac{\sqrt{mgk}\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} + \frac{m}{\sqrt{mgk}}C_y \\
y & = \frac{m\ln(kv_0\sin{\theta}\sin u + \sqrt{mgk}\cos u)}{k} + \frac{m}{\sqrt{mgk}}C_y \\
反向替换，u = \frac{t\sqrt{mgk}}{m} \\
\Rightarrow y & = \frac{m\ln(kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{t\sqrt{mgk}}{m}})}{k} + \frac{m}{\sqrt{mgk}}C_y
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_y\)：
\[\begin{aligned}
y_0 & = \frac{m\ln(kv_0\sin{\theta}\sin{\frac{0\cdot\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{0\cdot\sqrt{mgk}}{m}})}{k} + \frac{m}{\sqrt{mgk}}C_y \\
y_0 & = \frac{m\ln(\sqrt{mgk})}{k} + \frac{m}{\sqrt{mgk}}C_y \\
\frac{m}{\sqrt{mgk}}C_y & = y_0 -\frac{m\ln(\sqrt{mgk})}{k} \\
C_y & = \frac{\sqrt{mgk}}{m}y_0 -\frac{\sqrt{mgk}\ln(\sqrt{mgk})}{k} \\
\end{aligned}
\]
可得\(y\)：
\[\begin{aligned}
y & = \frac{m\ln(kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{t\sqrt{mgk}}{m}})}{k} - \frac{m\ln(\sqrt{mgk})}{k} + y_0 \\
y & = \frac{m\ln(kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{t\sqrt{mgk}}{m}}) - m\ln(\sqrt{mgk})}{k} + y_0 \\
y & = \frac{m\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{t\sqrt{mgk}}{m}}}{\sqrt{mgk}})}{k} + y_0 \\
y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} + \sqrt{mgk}\cos{\frac{t\sqrt{mgk}}{m}}}{\sqrt{mgk}}) + y_0 \\
\Rightarrow y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{t\sqrt{mgk}}{m}}) + y_0 \qquad \text{(6.1)}
\end{aligned}
\]
对于\(v_y\text{(5.2)}\)，有：
\[\begin{aligned}
v_y & = \frac{dy}{dt} \\
\sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})) & = \frac{dy}{dt}
\end{aligned}
\]
分离变量，两边同时积分得：
\[\begin{aligned}
dy & = \sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))dt \\
\int dy & = \int \sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))dt \\
\int dy & = -\sqrt{\frac{mg}{k}}\int \tan(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))dt \\
令 u = t\sqrt{\frac{gk}{m}} - & \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}), 则t = \frac{\sqrt{m}u + \sqrt{m}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})}{\sqrt{gk}}, dt = \frac{\sqrt{m}}{\sqrt{gk}}du \\
\int dy & = -\sqrt{\frac{mg}{k}}\int \tan(u)\frac{\sqrt{m}}{\sqrt{gk}}du \\
\int dy & = -\sqrt{\frac{m^2}{k^2}}\int \tan(u)du \\
\int dy & = -\frac{m}{k}\int \frac{\sin u}{\cos u}du \\
\int dy & = -\frac{m}{k}\int -\frac{1}{\cos u}d({\cos u}) \\
y + C_y & = \frac{m}{k}\ln|\cos u| \\
反向替换，u & = t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) \\
\Rightarrow y + C_y & = \frac{m}{k}\ln|\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))|
\end{aligned}
\]
注意到，当\(\, t = 0,\, x = x_0, y = y_0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)
那么可解得\(C_y\)：
\[\begin{aligned}
+ y_0C_y & = \frac{m}{k}\ln|\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| \\
C_y & = - + y_0\frac{m}{k}\ln|\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))|
\end{aligned}
\]
可得\(y\)：
\[\begin{aligned}
y + \frac{m}{k}\ln|\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| & = \frac{m}{k}\ln|\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| + y_0 \\
y & = \frac{m}{k}\ln|\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| - \frac{m}{k}\ln|\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| + y_0 \\
y & = \frac{m}{k}(\ln|\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))| -\ln|\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))|) + y_0 \\
\Rightarrow y & = \frac{m}{k}\ln|\frac{\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \qquad \text{(6.2)}\\
\end{aligned}
\]
轨迹方程
\[\begin{aligned}
x & = \frac{m}{k}\ln(\frac{m + kv_0\cos{\theta}t}{m}) + x_0\qquad \text{(4)} \\
y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{t\sqrt{mgk}}{m}}) + y_0 \qquad \text{(6.1)} \\
y & = \frac{m}{k}\ln|\frac{\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \qquad \text{(6.2)}
\end{aligned}
\]
由\(\text{(6.2)}\text{(7)}\)可得：
\[\begin{aligned}
x & = \frac{m}{k}\ln(\frac{m + kv_0\cos{\theta}t}{m}) + x_0 \\
\frac{k}{m}(x - x_0) & = \ln(\frac{m + kv_0\cos{\theta}t}{m}) \\
e^{\frac{k}{m}(x - x_0)} & = \frac{m + kv_0\cos{\theta}t}{m} \\
me^{\frac{k}{m}(x - x_0)} & = m + kv_0\cos{\theta}t \\
me^{\frac{k}{m}(x - x_0)} - m & = kv_0\cos{\theta}t \\
t & = \frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}} \qquad \text{(7)}\\
\end{aligned}
\]
联立\(\text{(6.1)}\text{(7)}\)或\(\text{(6.2)}\text{(7)}\)式，消去\(t\)，可得轨迹方程：
\[\begin{aligned}
y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{mgk}}{m}}) + y_0 \qquad \text{(8.1)} \\
y & = \frac{m}{k}\ln|\frac{\cos(\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \qquad \text{(8.2)}
\end{aligned}
\]
落回初始高度所需时间\(t\)
当\(v_y = 0\)，抛射体上升到最大高度\(y_{max}\)，由\(\text{(5.2)}\)可得：
\[\begin{aligned}
0 & = \sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})) \\
0 & = \tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})) \\
t\sqrt{\frac{gk}{m}} & = \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) \\
t & = \sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) \\
\end{aligned}
\]
即抛射体上升到最大高度所需时间\(t_m = \sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})\)
由\(\text{(6.2)}\)可得：
\[\begin{aligned}
y_{max} & = \frac{m}{k}\ln|\frac{\cos(t_m\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\sqrt{\frac{m}{gk}}\sqrt{\frac{gk}{m}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(0)}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
& 因为\arctan x \in (-\pi/2,\pi/2) \\
\Rightarrow y_{max} & = \frac{m}{k}\ln\frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))} + y_0 \\
\end{aligned}
\]
最大射程及其对应\(\theta\)
运用至英雄吊射或飞镖视觉
- 视觉识别装甲板或其他方法定位装甲板，得到装甲板坐标\((x_0, y_0)\)
- 云台枪口移至装甲板正中心
- 以此时云台pitch轴为y轴，枪口朝向为x轴建立坐标系
- 将装甲板坐标转换至该坐标系
- 已知初始射速\(v_0\)，目标装甲板坐标\((x_0, y_0)\)
对于\(\text{(4)}\)式，展开\(e^{-\frac{k}{m}t}\)，并取展开式的二次项：\[e^{-\frac{k}{m}t} \approx 1 - \frac{k}{m}t + \frac{1}{2}(\frac{k}{m})^2t^2 \\ x_0 = \frac{mv_0\cos{\theta}}{k}(1 - e^{- \frac{k}{m}t}) \\ \Rightarrow (\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2) = \frac{kx_0}{mv_0\cos{\theta}} \qquad \text{(i)}\\ \frac{1}{2}(\frac{k}{m})^2t^2 - \frac{k}{m}t + \frac{kx_0}{mv_0\cos{\theta}} = 0 \\ \frac{k^2}{2m^2}t^2 - \frac{k}{m}t + \frac{kx_0}{mv_0\cos{\theta}} = 0 \\ \frac{k}{2m}t^2 - t + \frac{x_0}{v_0\cos{\theta}} = 0 \\ t^2 - \frac{2m}{k}t + \frac{2mx_0}{kv_0\cos{\theta}} = 0 \\ \Rightarrow t = \frac{-(- \frac{2m}{k}) \pm \sqrt{(- \frac{2m}{k})^2 - 4\frac{2mx_0}{kv_0\cos{\theta}}}}{2} \\ t = \frac{\frac{2m}{k} \pm \sqrt{4(\frac{m}{k})^2 - 4\frac{2mx_0}{kv_0\cos{\theta}}}}{2} \\ t = \frac{\frac{2m}{k} \pm 2\sqrt{(\frac{m}{k})^2 - \frac{2mx_0}{kv_0\cos{\theta}}}}{2} \\ t = \frac{m}{k} \pm \sqrt{\frac{m}{k}(\frac{m}{k} - \frac{2x_0}{v_0\cos{\theta}})} \qquad \text{(ii)}\\ \]将\(\text{(i)}\text{(ii)}\)式，一并带入\(\text{(6)}\)式\[y_0 = \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2) - \frac{mg}{k}t \\ + y_0\frac{mg}{k}t = \frac{m}{k}(\frac{mg}{k} + v_0\sin{\theta})(\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2)\\ \frac{ + y_0\frac{mg}{k}t}{\frac{m}{k}(\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2)} = \frac{mg}{k} + v_0\sin{\theta}\\ \sin{\theta}= \frac{\frac{ + y_0\frac{mg}{k}t}{\frac{m}{k}(\frac{k}{m}t - \frac{1}{2}(\frac{k}{m})^2t^2)} - \frac{mg}{k}}{v_0} \\ \sin{\theta}= \frac{\frac{ + y_0\frac{mg}{k}t}{\frac{m}{k}\frac{kx_0}{mv_0\cos{\theta}}} - \frac{mg}{k}}{v_0} \\ \sin{\theta}= \frac{ + y_0\frac{mg}{k}t}{v_0 \cdot \frac{m}{k}\frac{kx_0}{mv_0\cos{\theta}}} - \frac{mg}{kv_0} \\ \sin{\theta}= \frac{ + y_0\frac{mg}{k}t}{v_0 \cdot \frac{m}{k}\frac{kx_0}{mv_0\cos{\theta}}} - \frac{mg}{kv_0} \\ \sin{\theta}= \frac{ + y_0\frac{mg}{k}t}{\frac{x_0}{\cos{\theta}}} - \frac{mg}{kv_0}\\ \Rightarrow \sin{\theta}= \frac{ + y_0\frac{mg}{k}(\frac{m}{k} \pm \sqrt{\frac{m}{k}(\frac{m}{k} - \frac{2x_0}{v_0\cos{\theta}})})}{\frac{x_0}{\cos{\theta}}} - \frac{mg}{kv_0}\\ \]最终得到只关于\(\theta\)的方程，用数值求解的方式，使用龙格-库塔(Runge-Kutta)法求得比较精确的解
这个解，即为解得pitch轴所需转动的角度
同时需要寻找一个角度，在弹道稳定的情况下，在初射速存在波动的情况下也能命中目标TODO
空气阻力与速度的二次方成正比
运动至最高点过程
由牛顿第二定律：
\[\begin{aligned}
ma_x & = - kv^2_x \\
ma_y & = - (mg + kv^2_y)
\end{aligned}
\]
也即：
\[\begin{aligned}
m\frac{dv_x}{dt} & = - kv^2_x & \text{(1)}\\
m\frac{dv_y}{dt} & = -mg - kv^2_y & \text{(2)} \\
\end{aligned} \\
\]
速度与位移方程
对\(\text{(1)}\text{(2)}\)式分离变量，且两边同时积分，
且当 \(t = 0,\, v_x = v_0\cos{\theta},\, v_y = v_0\sin{\theta}\)，
可得：
\[ v_x = \frac{mv_0\cos{\theta}}{m + ktv_0\cos{\theta}} \tag{3}
\]
\[ v_y = \frac{mg\tan(-\frac{t\sqrt{mgk}}{m}) + v_0\sin{\theta}\sqrt{mgk}}{\sqrt{mgk} - kv_0\sin{\theta}\tan(-\frac{t\sqrt{mgk}}{m})} \tag{4.1}
\]
\[ v_y = \sqrt{\frac{mg}{k}}\tan(-t\sqrt{\frac{gk}{m}} + \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})) \tag{4.2}
\]
对于\(v_x,v_y\)，有：
\[\begin{aligned}
v_x & = \frac{dx}{dt} \\
v_y & = \frac{dy}{dt}
\end{aligned}
\]
对\(\text{(3)}\text{(4)}\)式分离变量，且两边同时积分，
且当 \(t = 0,\, x = x_0,\, y = y_0\)，
可得：
\[ x = \frac{m}{k}\ln(1 + \frac{ktv_0\cos{\theta}}{m}) + x_0 \tag{5}
\]
\[ y = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{t\sqrt{mgk}}{m}}) + y_0 \tag{6.1}
\]
\[ y = \frac{m}{k}\ln|\frac{\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \tag{6.2}
\]
轨迹方程
\[\begin{aligned}
x & = \frac{m}{k}\ln(\frac{m + ktv_0\cos{\theta}}{m}) + x_0\qquad \text{(5)} \\
y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{t\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{t\sqrt{mgk}}{m}}) + y_0 \qquad \text{(6.1)} \\
y & = \frac{m}{k}\ln|\frac{\cos(t\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \qquad \text{(6.2)}
\end{aligned}
\]
由\(\text{(5)}\)可得：
\[\begin{aligned}
x & = \frac{m}{k}\ln(\frac{m + ktv_0\cos{\theta}}{m}) + x_0 \\
\frac{k}{m}(x - x_0) & = \ln(\frac{m + ktv_0\cos{\theta}}{m}) \\
e^{\frac{k}{m}(x - x_0)} & = \frac{m + ktv_0\cos{\theta}}{m} \\
me^{\frac{k}{m}(x - x_0)} & = m + ktv_0\cos{\theta} \\
me^{\frac{k}{m}(x - x_0)} - m & = ktv_0\cos{\theta} \\
t & = \frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}} \qquad \text{(7)}\\
e^{\frac{k}{m}(x - x_0)} & = 1 + \frac{ktv_0\cos{\theta}}{m} \\
e^{\frac{k}{m}(x - x_0)} - 1 & = \frac{ktv_0\cos{\theta}}{m} \\
m(e^{\frac{k}{m}(x - x_0)} - 1) & = ktv_0\cos{\theta} \\
t & = \frac{m}{kv_0\cos{\theta}}(e^{\frac{k}{m}(x - x_0)} - 1) \\
\end{aligned}
\]
联立\(\text{(6.1)}\text{(7)}\)或\(\text{(6.2)}\text{(7)}\)式，消去\(t\)，可得轨迹方程：
\[\begin{aligned}
y & = \frac{m}{k}\ln(\frac{kv_0\sin{\theta}\sin{\frac{\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{mgk}}{m}} }{\sqrt{mgk}} + \cos{\frac{\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{mgk}}{m}}) + y_0 \qquad \text{(8.1)} \\
y & = \frac{m}{k}\ln|\frac{\cos(\frac{me^{\frac{k}{m}(x - x_0)} - m}{kv_0\cos{\theta}}\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \qquad \text{(8.2)}
\end{aligned}
\]
当\(v_y = 0\)，抛射体上升到最大高度\(y_{max}\)，由\(\text{(4.2)}\)可得：
\[ t = \sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})
\]
即抛射体上升到最大高度所需时间
\[ t_m = \sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) \tag{9}
\]
由\(\text{(6.2)}\)可得：
\[\begin{aligned}
y_{max} & = \frac{m}{k}\ln|\frac{\cos(t_m\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\sqrt{\frac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})\sqrt{\frac{gk}{m}} - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\sqrt{\frac{m}{gk}}\sqrt{\frac{gk}{m}}\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}) - \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{\cos(0)}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
y_{max} & = \frac{m}{k}\ln|\frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))}| + y_0 \\
& 因为\arctan x \in (-\pi/2,\pi/2) \\
y_{max} & = \frac{m}{k}\ln\frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))} + y_0 \\
\end{aligned}
\]
令\(c = \arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}})\)，那么\(\tan c = v_0\sin{\theta}\sqrt{\frac{k}{mg}}\)
\[ 1 + \tan^2 c = 1 + (v_0\sin{\theta}\sqrt{\frac{k}{mg}})^2 = \frac{1}{\cos^2c} \\
\Rightarrow \frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))} = \sqrt{1 + (v_0\sin{\theta}\sqrt{\frac{k}{mg}})^2}
\]
\[\begin{aligned}
y_{max} & = \frac{m}{k}\ln\frac{1}{\cos(\arctan(v_0\sin{\theta}\sqrt{\frac{k}{mg}}))} + y_0 \\
\Rightarrow y_{max} & = \frac{m}{k}\ln\sqrt{1 + (v_0\sin{\theta}\sqrt{\frac{k}{mg}})^2} + y_0
\end{aligned}
\]
由最高点下降过程
\(v_{y_2} < 0\)
由牛顿第二定律：
\[\begin{aligned}
ma_{x_2} & = - kv^2_{x_2} \\
ma_{y_2} & = - mg + kv^2_{y_2}
\end{aligned}
\]
也即：
\[\begin{aligned}
m\frac{dv_{x_2}}{dt} & = - kv^2_{x_2} & \text{(1)}\\
m\frac{dv_{y_2}}{dt} & = -mg + kv^2_{y_2} & \text{(2)} \\
\end{aligned} \\
\]
在最高点时，由\((3)(9)\)式，可得抛射体速度：
\[ v_{0x_2} = \cfrac{mv_0\cos{\theta}}{m + k\sqrt{\cfrac{m}{gk}}\arctan(v_0\sin{\theta}\sqrt{\cfrac{k}{mg}})v_0\cos{\theta}} \\
v_{0y_2} = 0
\]
速度与位移方程
对\(\text{(1)}\text{(2)}\)式分离变量，且两边同时积分，
且当 \(t = t_m,\, v_{x_2} = v_{0x_2},\, v_{y_2} = 0\)，
可得：
\[ v_{x_2} = \frac{mv_0\cos{\theta}}{m + ktv_0\cos{\theta}} \tag{3}
\]
\[ \frac{dv_{y_2}}{dt} = -g + \frac{k}{m}v^2_{y_2} \\
\frac{dv_{y_2}}{-g + \frac{k}{m}v^2_{y_2} } = dt \\
\frac{1}{2}\sqrt{\frac{m}{gk}}\ln\frac{|kv_{y_2} - \sqrt{mgk}|}{kv_{y_2} + \sqrt{mgk}} + C_{v_{y_2}} = t \\
C_{v_{y_2}} = t_m \\
\frac{1}{2}\sqrt{\frac{m}{gk}}\ln\frac{|kv_{y_2} - \sqrt{mgk}|}{kv_{y_2} + \sqrt{mgk}} + t_m = t \\
\frac{1}{2}\sqrt{\frac{m}{gk}}\ln\frac{|kv_{y_2} - \sqrt{mgk}|}{kv_{y_2} + \sqrt{mgk}} = t - t_m\\
\ln\frac{|kv_{y_2} - \sqrt{mgk}|}{kv_{y_2} + \sqrt{mgk}} = 2\sqrt{\frac{gk}{m}}(t - t_m)\\
\frac{|kv_{y_2} - \sqrt{mgk}|}{kv_{y_2} + \sqrt{mgk}} = e^{2\sqrt{\frac{gk}{m}}(t - t_m)}\\
|kv_{y_2} - \sqrt{mgk}| = e^{2\sqrt{\frac{gk}{m}}(t - t_m)}(kv_{y_2} + \sqrt{mgk})\\
-kv_{y_2} + \sqrt{mgk} = e^{2\sqrt{\frac{gk}{m}}(t - t_m)}kv_{y_2} + e^{2\sqrt{\frac{gk}{m}}(t - t_m)}\sqrt{mgk}\\
-kv_{y_2} - e^{2\sqrt{\frac{gk}{m}}(t - t_m)}kv_{y_2} = e^{2\sqrt{\frac{gk}{m}}(t - t_m)}\sqrt{mgk} - \sqrt{mgk}\\
-kv_{y_2} = \frac{\sqrt{mgk}(e^{2\sqrt{\frac{gk}{m}}(t - t_m)} - 1)}{1 + e^{2\sqrt{\frac{gk}{m}}(t - t_m)}}\\
v_{y_2} = \frac{\sqrt{mgk}(1 - e^{2\sqrt{\frac{gk}{m}}(t - t_m)})}{k(1 + e^{2\sqrt{\frac{gk}{m}}(t - t_m)})}\\
v_y = -\sqrt{\frac{mg}{k}}\th{(\sqrt{\frac{k}{mg}}g(t-t_m))}
\]
对于\(v_x,v_y\)，有：
\[\begin{aligned}
v_x & = \frac{dx}{dt} \\
v_y & = \frac{dy}{dt}
\end{aligned}
\]
对\(\text{(3)}\text{(4)}\)式分离变量，且两边同时积分，
且当 \(t = t_m,\, x = x_m,\, y = y_m\)，
可得：
\[ x = \frac{m}{k}\ln(1 + \frac{ktv_0\cos{\theta}}{m}) + x_0 \tag{5}
\]
\[ v_y = -\sqrt{\frac{mg}{k}}\tanh{(\sqrt{\frac{k}{mg}}g(t-t_m))} \\
\frac{dy}{dt} = -\sqrt{\frac{mg}{k}}\tanh{(\sqrt{\frac{k}{mg}}g(t-t_m))} \\
dy = -\sqrt{\frac{mg}{k}}\tanh{(\sqrt{\frac{k}{mg}}g(t-t_m))}dt \\
\int dy = \int -\sqrt{\frac{mg}{k}}\tanh{(\sqrt{\frac{k}{mg}}g(t-t_m))}dt \\
\int dy = -\sqrt{\frac{mg}{k}} \int \tanh{(\sqrt{\frac{k}{mg}}g(t-t_m))}dt \\
y + C_y = -\frac{m}{k}\ln\coth(\sqrt{\frac{k}{mg}}g(t-t_m))\\
y_m + C_y = -\frac{m}{k}\ln\coth(\sqrt{\frac{k}{mg}}g(t_m-t_m)) \\
C_y = -y_m \\
y = y_m -\frac{m}{k}\ln\coth(\sqrt{\frac{k}{mg}}g(t-t_m))\\
\]
消去t
\[ x = \frac{m}{k}\ln(1 + \frac{ktv_0\cos{\theta}}{m}) + x_0 \\
y = y_m -\frac{m}{k}\ln\coth(\sqrt{\frac{k}{mg}}g(t-t_m))\\
t = \frac{m}{kv_0\cos{\theta}}(e^{\frac{k}{m}(x - x_0)} - 1) \\
y = y_m -\frac{m}{k}\ln\coth(\sqrt{\frac{k}{mg}}g(\frac{m}{kv_0\cos{\theta}}(e^{\frac{k}{m}(x - x_0)} - 1)-t_m))\\
\]
