Exercise_05:The Trajectory of a Cannon Shell

Abstract

• use python to do homework 2.6

Background

• Use the Euler method to calculate cannon shell trajectories,ignoring both air drag and the effect of air density.
  $$\frac{d^2x}{dt^2}=0 ,\frac{d^2y}{dt^2}=-g$$

Main body

Analysis:

$$\frac{dx}{dt}=v_x , \frac{dv_x}{dt}=0$$
$$\frac{dy}{dt}=v_y , \frac{dv_y}{dt}=-g$$
so: $$x_{i+1}=x_i+v_{x,i}\Delta t$$
$$v_{i+1}=v_{x,i}$$
$$y_{x,i+1}=y_i+v_{y,i}\Delta t$$
$$v_{y,i+1}=v_{y,i}-g \Delta t$$
This is primary code：

import pylab as pl
from math import *
class cannon:
    def __init__(self,init_v=10,theta =60,
                 init_x=0,init_y=0,time_step=0.01 ):     
        self.vx = [init_v*cos(theta*2*pi/360)]
        self.vy = [init_v*sin(theta*2*pi/360)]
        self.x = [init_x]
        self.y = [init_y]
        self.t = [0]
        self.B_m = B_m
        self.dt = time_step        
    def run(self):
        while self.y[-1]>=0:
            self.x.append(self.x[-1] + self.dt * self.vx[-1])
            self.y.append(self.y[-1] + self.dt * self.vy[-1])
            self.vy.append(self.vy[-1] - self.dt * 9.8 )
    def show_results(self):
        pl.plot(self.x, self.y,color="black",
                linestyle="-",label="60")
        pl.legend(loc='upper right')
        pl.xlabel('x(km)')
        pl.ylabel('y(km)')
        pl.ylim(0,4)
        pl.show()
a = cannon()
a.run()
a.show_results()

And this is a picture, which shows three different launch angle.

Conclusion

• Ignoring the air drag, we can easy see that when launch angle 30 degree and 60 degree reach the same distance , and when launch angle are 45 degree, the distance are the furthest. The theoretical result is identical with the picture.

• After two exercise, I understand how to use python and Euler method to caltulate some appoxiation.

Acknowledgement

Thanks for teacher cai's PPT.