Tuesday, August 13, 2013

Applications of Higher Order Differential Equations: Missile Trajectory


One of the most powerful applications of  higher order differential equations is to the ballistic trajectory of a missile, for which a diagram of the type shown is often used, say to define the launch angle qo, as well as the final x, y- coordinates (e.g. at impact). Consider then the following problem:

Find the  x- and y-coordinates of the points on the trajectory of a missile launched at an angle of 80 degrees with an initial velocity of 100,000 f/s if the air resistance is 0.01mv. Find the value of x and y after 10 seconds.

Data:    q = 80o,   k = 0.01mv   v = 105 fs-1 and t = 10s

The differential equation can be written:

m(d2x/dt2) =  - k (dx/dt)/ v = - (0.01mv)x’/ v  = - 0.01mx’

Similarly:

my” = -mg – (0.01mv)y’/v  =   -mg – 0.01my’

 
Yielding the simultaneous pair:

x” + 0.01x’ = 0

y” + 0.01 y’ = -g

 
We know:

x’o = vo cos (q) = vo cos (80) = (105) cos 80

y’o = vo sin (q) = vo sin (80) = (105) sin 80

 
General Solution:

x = c1 + c2 (e -0.01t)

y = c3 + c4((e -0.01t)  - 100 gt

x’ = -0.01c2 (e -0.01t)

y’ = -0.01c4((e -0.01t)  - 100 g

 
At t = 0, x = 0 so:

0 =  c1 + c2 (e -0.01t)

 
But:

x’o  = vo cos (80) = (105) cos 80 =  -0. 01c2 (e -0.01t)

Then: c2 = - (107) cos 80

So: 0 = c1 + ( - (107) cos 80) or c1 =  (107) cos 80

 
At t = 0, y = 0:

Then: 0 = c3 + c4 (e -0.01t)

And: y’o  = vo sin (80) = (105) sin 80 = -0.01c4(e -0.01t)  - 100 g

So: c4 = -(107) sin 80 – 10000g

\ c3 = - c4 (e -0.01t) = 107 sin 80 + 10000g


 
To obtain the x and y-coordinates:

First, the x-coordinate:

x = c1 + c2 (e -0.01t) = 107 cos 80 – (107 cos 80)( e -0.01t)

x = 107 cos 80 [1 - ( e -0.01t)] =  107 (0.17365) [1 - ( e -0.01t)]

 
or:   x = 1736500 [1 - ( e -0.01t)]
 

 
The y-coordinate:

y = c3 + c4 ( e -0.01t)

Or:  y = 107 sin 80 + 10000g – (107 sin 80 + 10000g(e -0.01t))

y = 107 sin 80 + 10000g(1 - e -0.01t) – 100 gt

y = [107 (0.9848) + 320,000](1 - e -0.01t) – 100 gt

 
y = 1016800(1 - e -0.01t) – 100 gt

After 10 seconds:

x = 1736500 [1 - ( e -0.01t)] = x = 1736500 [1 - ( e -0.01(10))]

 
x = 165, 200 ft. (» 31. 3 miles)

y = 1016800(1 - e -0.01(10)) – 100(32)(10)

y =  930,580 ft.  ( » 177 miles)


Problem for the Math Maven:

Re-work the problem but obtain the coordinates in kilometers, instead of miles. (Hint: you will use an initial velocity of  30,300 m/s instead of 100,000 f/s. Also you will use g = 10 ms-2   instead of g = 32 fs-2) What differences did you find in your final values - from those in the blog problem - and how might you account for them? (Convert  either miles to km, or vice versa, to compare).

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