Exercise_06: Chapter 2 problem 2.9 & 2.10: The cannon

一.摘要：

Generalize the program developed for the previous problem so that it can deal with situations in which the target is at a different altitude than the cannon.

二.背景介绍（公式推导）：

1.From Newton's second law: $$\ddot{x}=0,$$ $$\ddot{y}=-g,$$

2..Considering the air drag and assuming the air density is constant: $$F_{drag}=-B_2*v^2,①$$ $$F_{drag,x}=-B_2*v*v_x,$$ $$F_{drag,y}=-B_2*v*v_y,$$

3.Including the air drag,the reduced air density at high altitudes and $v_{wind}$: $$\rho=\rho_0*(1-\frac{ay}{T_0})^{\alpha},(the adiabatic progress)$$ $$F'_{drag}=\frac{\rho}{\rho_0}*F'_{drag}(y=0)$$ $$F_{drag,x}=-B_2\sqrt{v^2+(v_{wind})^2-2v_x*v_{wind}}*(v_x-v_{wind})$$ $$F_{drag,y}=-B_2\sqrt{v^2+(v_{wind})^2-2v_x*v_{wind}}*v_y$$

so, $$x_{i+1}=x_i+v_{x,i}*{\Delta}t$$ $$v_{x,i+1}=v_{x,i}+\frac{F_{drag,x}}{m}*{\Delta}t$$ $$y_{i+1}=x_i+v_{y,i}*{\Delta}t$$ $$v_{y,i+1}=v_{y,i}-g*{\Delta}t+\frac{F_{drag,y}}{m}*{\Delta}t$$

三.正文:

1.Results of the program including the air drag,the reduced air density at high altitudes and $v_{wind}$:

代码1

2.A defined target is at different $\theta$s:

Code2

3.A defined target is at different altitudes:

Code3

四结论：

1.当v=700m/s时,$\theta$=40和55都会经过the target.

2.当$\theta$=55时，初始速度v0越大，炮弹运动轨迹的最高点对应的x越大；当v0=700m/s时，炮弹运动轨迹会经过点（14516.1，5248.95）。