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 q_o, 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 = 80^o,   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 = v_o cos (q) = v_o cos (80) = (10⁵) cos 80

y’_o = v_o sin (q) = v_o 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  = v_o 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  = v_o 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).