Trajectory of a Cannon Shell

程熹 2013301020038 物基一班
Homework

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Abstract

This article discusses the problem what's a cannon shell trajectory at different initial velocity and different angle.By changing the initial velocity and angle, we can find what's the angle that a cannon shell can move in the longest distance. At later discussion, we include in air resistance to find its influence.

Key Word

Cannon shell movement,projectile motion,euler method.

Introduction

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth's surface, and it moves along a curved path under the action of gravity only. The implication here is that air resistance is negligible, or in any case is being neglected in all of these equations. The only force of significance that acts on the object is gravity, which acts downward to cause a downward acceleration. Because of the object's inertia, no external horizontal force is needed to maintain the horizontal motion.

Main Features

Let the projectile be launched with an initial velocity v0 . which can be expressed as the sum of horizontal and vertical components as follows:

$${\displaystyle \mathbf {v} _{0}=v_{0x}\mathbf {i} +v_{0y}\mathbf {j} }$$ .
The components V0xand V0ycan be found if the initial launch angle, {\displaystyle \theta } , is known:
$${\displaystyle v_{0x}=v_{0}\cos \theta }$$ $${\displaystyle v_{0y}=v_{0}\sin \theta }$$ .
The total time {\displaystyle t} for which the projectile remains in the air is called the time of flight.

$${\displaystyle y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$$
After the flight, the projectile returns to the horizontal axis, so y=0
$${\displaystyle 0=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$$ $${\displaystyle v_{0}t\sin(\theta )={\frac {1}{2}}gt^{2}}$$ $${\displaystyle v_{0}\sin(\theta )={\frac {1}{2}}gt}$$ $${\displaystyle t={\frac {2v_{0}\sin(\theta )}{g}}}$$
Note that we have neglected air resistance on the projectile.

Result

1.Initial velocity=100, initial angle=30

2.Initial velocity=100,initial angle=45

3.Initial velocity=100,initial angle=60

4.Initial velocity=100

5.Initial velocity=500

Conclusion

As we know, if there won't exist air drag and the altitude, at the angle of 45, the cannon shell can move at the longest distance. However, from the upper diagram we can find, at the low speed ,the air drag and the altitude will have little influence on the largest angle, as we can find, at V0=100, the angle that can make the longest distance is arond 45 diagree. However, on the fourth diagram, the angle around 55 can make the cannon shell travel longest. So I believe that the air drag and the altitude will lead to a larger initial angle for the cannon shell to travel longest.

Acknowledgement

This program is finished by myself.

Reference

Computational Physics(second edition) Nicholas J.Giordano and Hisao Nakanishi
Wikipedia Projectile motion
https://en.wikipedia.org/wiki/Projectile_motion