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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