Example 4.  An arrow is shot upward from the origin with an initial velocity of  300 ft/sec.
Assume that air resistance is proportional to the [Graphics:Images/ProjectileMotionMod_gr_284.gif]power of the velocity,  [Graphics:Images/ProjectileMotionMod_gr_285.gif],   and use the model  
        [Graphics:Images/ProjectileMotionMod_gr_286.gif].  
Find the velocity and position as a function of time, and plot the position function.

Solution 4.

Compute the solution using the Runge-Kutta method for second order D.E.'s.

[Graphics:../Images/ProjectileMotionMod_gr_287.gif]



[Graphics:../Images/ProjectileMotionMod_gr_288.gif]
[Graphics:../Images/ProjectileMotionMod_gr_289.gif]
[Graphics:../Images/ProjectileMotionMod_gr_290.gif]
[Graphics:../Images/ProjectileMotionMod_gr_291.gif]
[Graphics:../Images/ProjectileMotionMod_gr_292.gif]
[Graphics:../Images/ProjectileMotionMod_gr_293.gif]
[Graphics:../Images/ProjectileMotionMod_gr_294.gif]
[Graphics:../Images/ProjectileMotionMod_gr_295.gif]
[Graphics:../Images/ProjectileMotionMod_gr_296.gif]
[Graphics:../Images/ProjectileMotionMod_gr_297.gif]
[Graphics:../Images/ProjectileMotionMod_gr_298.gif]

The solution we seek is the first coordinate in the 2D system.

[Graphics:../Images/ProjectileMotionMod_gr_299.gif]

Now we can plot the solution.

[Graphics:../Images/ProjectileMotionMod_gr_300.gif]


[Graphics:../Images/ProjectileMotionMod_gr_301.gif]

[Graphics:../Images/ProjectileMotionMod_gr_302.gif]
[Graphics:../Images/ProjectileMotionMod_gr_303.gif]
[Graphics:../Images/ProjectileMotionMod_gr_304.gif]
[Graphics:../Images/ProjectileMotionMod_gr_305.gif]
[Graphics:../Images/ProjectileMotionMod_gr_306.gif]

The graph should lie between the ones obtained in examples 2 and 3.
Let's see if this happens.

[Graphics:../Images/ProjectileMotionMod_gr_307.gif]


[Graphics:../Images/ProjectileMotionMod_gr_308.gif]

The graph does lie where it should lie.

 

 

We are done.  

Aside.  We can "try" to get an analytic solution.

[Graphics:../Images/ProjectileMotionMod_gr_309.gif]



[Graphics:../Images/ProjectileMotionMod_gr_310.gif]
[Graphics:../Images/ProjectileMotionMod_gr_311.gif]
[Graphics:../Images/ProjectileMotionMod_gr_312.gif]
[Graphics:../Images/ProjectileMotionMod_gr_313.gif]
[Graphics:../Images/ProjectileMotionMod_gr_314.gif]


Clearly, Mathematica is having a difficult time of it.
If the initial condition is not used then Mathematica can go a little further in the solution.

[Graphics:../Images/ProjectileMotionMod_gr_315.gif]



[Graphics:../Images/ProjectileMotionMod_gr_316.gif]
[Graphics:../Images/ProjectileMotionMod_gr_317.gif]

We should trust our Runge-Kutta solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004