Homework 06

Question 01

（1）上升阶段受力分析可得：

\[ \begin{gather} -mg-f(v)=ma=m\frac{dv}{dt}\\ \therefore \frac{dv}{dt}=-g-\frac{f(v)}{m}\\ \Rightarrow -\frac{dv}{g+\frac{f(v)}{m}}=dt \end{gather} \]

两边积分可得 \(t_a=\int_0^{v_0}\frac{mdv}{mg+f(v)}\)

下降阶段同理受力分析得 \(t_d=\int_0^{v_f}\frac{mdv}{mg-f(v)}\)

考虑上升距离：

\[ \begin{gather} (mg+f(v))dx=\frac{1}{2}m[(v-dv)^2-v^2]\\ \Rightarrow -mvdv=(mg+f(v))dx \end{gather} \]

两边积分可得 \(h=\int_0^{v_0}\frac{mvdv}{mg+f(v)}\)

下降阶段同理可得 \(h=\int_0^{v_f}\frac{mvdv}{mg-f(v)}\)

由上升下降距离相同有 \(\int_0^{v_0}\frac{mvdv}{mg+f(v)}=\int_0^{v_f}\frac{mvdv}{mg-f(v)}\)

（2）由 \(mg+f(v)>mg-f(v)\)，所以 \(v_0>v_f,t_a<t_d\)

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Question 02

（1）易得在空中的时间为 \(t=\frac{2v\sin\theta}{g}\)，水平速度为 \(v_x=v\cos\theta\)，竖直速度为 \(v_y=v\sin\theta-gt\)

所以长度为 \(L(v,\theta)=2\int_0^{\frac{v\sin\theta}{g}}\sqrt{(v\cos\theta)^2+(v\sin\theta-gt)^2}dt\)

令 \(s=\sqrt{(v\cos\theta)^2+(v\sin\theta-gt)^2}\)

则 \(L(v,\theta)=(gt-v\sin\theta)s+(\frac{v\cos\theta}{g})^2\ln[(gt-v\sin\theta)+s]|_0^{\frac{v\sin\theta}{g}}=\frac{v^2\sin\theta}{g}+\frac{v^2\cos^2\theta}{g}\ln\frac{\cos\theta}{1-\sin\theta}\)

（2）\(L(v,\frac{\pi}{4})=\frac{\sqrt{2}v^2}{2g}+\frac{v^2}{2g}\ln\frac{\sqrt{2}}{2-\sqrt{2}}\approx 1.1\frac{v^2}{g}\)

\(L(v,\frac{\pi}{2})=\frac{v^2}{g}\)

\(\therefore L(v,\frac{\pi}{4})>L(v,\frac{\pi}{2})\)

（3）\(\frac{\partial L}{\partial\theta}=\frac{v^2\cos\theta}{g}-\frac{2v^2\sin\theta\cos\theta}{g}\ln\frac{\cos\theta}{1-\sin\theta}+\frac{v^2\cos\theta}{g}=0\)

\(\therefore\sin\theta\ln\frac{\cos\theta}{1-\sin\theta}=1\)

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