背景知识#
竖直上抛运动：指物体以某一初速度竖直向上抛出 (不考虑空气阻力), 只在重力作用下所做的运动.
若研究在竖直上抛运动中，一小球所能达到的最高位置的距离 $ H $ 时，则有一般思路:
Copy 若初速度为 v 0 , 重力加速度为 g , 则有 H = v 0 2 2 g 若初速度为 v_0, 重力加速度为 g, 则有 H = \frac{v_0^2}{2g} 若初速度为 v 0 , 重力加速度为 g , 则有 H = 2 g v 0 2 那么，如果将这个问题改变一下，若考虑空气阻力对竖直上抛运动的影响，那么结果会是如何呢？
已知条件#
已知空气阻力的关系式 $ f = \frac {1}{2} c \rho S v^2 $, 其中 $ C $ 为空气阻力系数，$ \rho $ 为空气密度，$ S $ 物体迎风面积，$ v $ 为物体与空气的相对运动速度，小球的质量为 $ m $, 重力加速度为 $ g $.
设 $ H $ 为小球所能达到的最高位置的距离.
不考虑 $ c, \rho, S $ 对本题目的影响，且小球直径 $ d \ll H $.
设水平向下为小球运动的正方向，根据牛顿第二定律 $ F = ma $, 则有 $ ma = mg + \frac {1}{2} c \rho S v^2 $,
令 $ \lambda = \frac {1}{2} c \rho S $, 即 $ ma = mg + \lambda v^2 $,
Copy ∵ a = v ˙ = d v d t ∴ m d v d t = m g + λ v 2 ⟹ d v d t = g + λ m v 2 \begin{align}
\because a &= \dot{v} = \frac{dv}{dt} \\
\therefore m \frac{dv}{dt} &= mg + \lambda v^2 \\
\implies \frac{dv}{dt} &= g + \frac{\lambda}{m} v^2 \\
\end{align} ∵ a ∴ m d t d v ⟹ d t d v = v ˙ = d t d v = m g + λ v 2 = g + m λ v 2 令 $ \frac {\lambda}{m} = \mu $,
Copy ∴ d v d t = g + μ v 2 ⟹ d v g + μ v 2 = d t ⟹ ∫ d v g + μ v 2 = ∫ d t + C ⟹ 1 g ∫ d v 1 + μ g v 2 = t + C ⟹ 1 g ∫ g μ d ( μ g v ) 1 + ( μ g v ) 2 = t + C ⟹ 1 g g μ arctan ( μ g v ) = t + C ⟹ 1 μ g arctan ( μ g v ) = t + C ( ∗ ) \begin{align}
\therefore \frac{dv}{dt} &= g + \mu v^2 \\
\implies \frac{dv}{g + \mu v^2} &= dt \\
\implies \int
\frac{dv}{g + \mu v^2}
&=
\int
dt + C \\
\implies \frac{1}{g} \int
\frac{dv}{1 + \frac{\mu}{g}v^2}
&=
t + C \\
\implies \frac{1}{g} \int
\frac{\frac{\sqrt{g}}{\sqrt{\mu}}d(\frac{\sqrt{\mu}}{\sqrt{g}}v)}{1 + (\frac{\sqrt{\mu}}{\sqrt{g}}v)^2}
&=
t + C \\
\implies \frac{1}{g}\frac{\sqrt{g}}{\sqrt{\mu}}\arctan(\sqrt{\frac{\mu}{g}}v) &= t + C \\
\implies \sqrt{\frac{1}{\mu g}}\arctan(\sqrt{\frac{\mu}{g}}v) &= t + C (*)
\end{align} ∴ d t d v ⟹ g + μ v 2 d v ⟹ ∫ g + μ v 2 d v ⟹ g 1 ∫ 1 + g μ v 2 d v ⟹ g 1 ∫ 1 + ( g μ v ) 2 μ g d ( g μ v ) ⟹ g 1 μ g arctan ( g μ v ) ⟹ μg 1 arctan ( g μ v ) = g + μ v 2 = d t = ∫ d t + C = t + C = t + C = t + C = t + C ( ∗ ) 当 $ t = 0 $ 时，$ v = -v_0 $, 有:
Copy C = − 1 μ g arctan ( μ g v 0 ) ( ∗ ∗ ) \begin{align}
C &= -\sqrt{\frac{1}{\mu g}}\arctan(\sqrt{\frac{\mu}{g}}v_0) (**)
\end{align} C = − μg 1 arctan ( g μ v 0 ) ( ∗ ∗ ) 当 $ v = 0 $ 时，有:
Copy t = 1 μ g arctan ( μ g v 0 ) \begin{align}
t &= \sqrt{\frac{1}{\mu g}}\arctan(\sqrt{\frac{\mu}{g}}v_0)
\end{align} t = μg 1 arctan ( g μ v 0 ) 若欲将 $ (*) $ 转换为 $ v $ 与 $ t $ 之间的显式表达式，则需:
Copy 将 ( ∗ ) 变换 , 得 : arctan ( μ g v ) = μ g ( t + C ) ⟹ μ g v = tan [ μ g ( t + C ) ] ⟹ v = g μ tan [ μ g ( t + C ) ] \begin{align}
将 (*) 变换, 得: \arctan(\sqrt{\frac{\mu}{g}}v) &= \sqrt{\mu g} (t + C) \\
\implies \sqrt{\frac{\mu}{g}}v &= \tan[\sqrt{\mu g} (t + C)] \\
\implies v &= \sqrt{\frac{g}{\mu}}\tan[\sqrt{\mu g} (t + C)]
\end{align} 将 ( ∗ ) 变换 , 得 : arctan ( g μ v ) ⟹ g μ v ⟹ v = μg ( t + C ) = tan [ μg ( t + C )] = μ g tan [ μg ( t + C )] 又 $ \because v = \dot {x} = \frac {dx}{dt} $, 有:
Copy d x d t = g μ tan [ μ g ( t + C ) ] ⟹ ∫ d x d t d t = g μ ∫ tan [ μ g ( t + C ) ] d t + C ′ ⟹ x = g μ ∫ sin [ μ g ( t + C ) ] cos [ μ g ( t + C ) ] d t + C ′ ⟹ x = − g μ ∫ d c o s [ μ g ( t + C ) ] μ g ( t + C ) + C ′ ⟹ x = − 1 μ ln ∣ cos [ μ g ( t + C ) ] ∣ + C ′ \begin{align}
\frac{dx}{dt} &= \sqrt{\frac{g}{\mu}}\tan[\sqrt{\mu g} (t + C)] \\
\implies \int
\frac{dx}{dt}dt &=
\sqrt{\frac{g}{\mu}} \int
\tan[\sqrt{\mu g} (t + C)]dt + C' \\
\implies x &= \sqrt{\frac{g}{\mu}} \int
\frac{\sin[\sqrt{\mu g} (t + C)]}{\cos[\sqrt{\mu g} (t + C)]}dt + C' \\
\implies x &= -\sqrt{\frac{g}{\mu}} \int
\frac{dcos[\sqrt{\mu g} (t + C)]}{\sqrt{\mu g} (t + C)} + C' \\
\implies x &= -\frac{1}{\mu} \ln{|\cos[\sqrt{\mu g}(t + C)]|} + C'
\end{align} d t d x ⟹ ∫ d t d x d t ⟹ x ⟹ x ⟹ x = μ g tan [ μg ( t + C )] = μ g ∫ tan [ μg ( t + C )] d t + C ′ = μ g ∫ cos [ μg ( t + C )] sin [ μg ( t + C )] d t + C ′ = − μ g ∫ μg ( t + C ) d cos [ μg ( t + C )] + C ′ = − μ 1 ln ∣ cos [ μg ( t + C )] ∣ + C ′ 由 $ H = x_{|t = \sqrt {\frac {1}{\mu g}}\arctan (\sqrt {\frac {\mu}{g}} v_0)} - x_{|t = 0}$,
注意到，其中 $ t + C = 0 $, 则，
Copy H = 0 − C ′ − [ − 1 μ ln ∣ cos ( C μ g ) ∣ + C ′ ] ⟹ H = 1 μ ln ∣ cos ( C μ g ) ∣ + C ′ ( ∗ ∗ ∗ ) \begin{align}
H = 0 - C' - [-\frac{1}{\mu} \ln{|\cos(C\sqrt{\mu g})|} + C'] \\
\implies H = \frac{1}{\mu} \ln{|\cos(C\sqrt{\mu g})|} + C' (***)
\end{align} H = 0 − C ′ − [ − μ 1 ln ∣ cos ( C μg ) ∣ + C ′ ] ⟹ H = μ 1 ln ∣ cos ( C μg ) ∣ + C ′ ( ∗ ∗ ∗ ) 将 $ (), \lambda = \mu m = \frac {1}{2} c \rho S v^2 $, 代入 $ ( *) $ 并整理，得:
Copy H = 2 m c ρ S ln ∣ cos [ arctan ( c ρ S v 0 2 m g ) ] ∣ \begin{align}
H &= \frac{2m}{c \rho S}\ln{|\cos[\arctan(\frac{c \rho S v_0}{2mg})]|}
\end{align} H = c ρS 2 m ln ∣ cos [ arctan ( 2 m g c ρS v 0 )] ∣ 上式即为所求.
此文由 Mix Space 同步更新至 xLoghttps://ms.rescueme.life/posts/default/vertical-upward-throwing-motion-considering-air-resistance