*Find the x- and y-coordinates of the points on the trajectory of a rocket shot 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.*

We have:

Θ = 80 deg, k = 0.01mv, v = 10^5 m/s and t = 10s

The differential equation can be written:

m(d^2x/dt^2) = - k (dx/dt)/ v = - (0.01mv)x’/ v = - 0.01mx’

Similarly:

my” = -mg – (0.01mv)y’/v

where as usual y" denotes the 2nd derivative wrt time, and y' the 1st derivative.

Whence:

my" = = -mg – 0.01my’

Yielding the simultaneous pair

:

x” + 0.01x’ = 0

y” + 0.01 y’ = -g

We know:

x’_o = v_o cos (Θ) = v_o cos (80) = (10^5) cos 80

y’_o = v_o sin (Θ) = v_o sin (80) = (10^5) sin 80

General Solutions are:

x = c1 + c2 exp(-0.01t)

y = c3 + c4(exp( -0.01t) - 100 gt

x’ = -0.01c2 exp(-0.01t)

y’ = -0.01c4 exp(-0.01t) - 100 g

For initial conditions, at t = 0, x = 0 so:

0 = c1 + c2 exp(-0.01t)

But:

x’_o = v_o cos (80) = (10^5) cos 80 =

-0. 01c2 exp(-0.01t)

Then: c2 = - (10^7) cos 80

So: 0 = c1 + (- (10^7) cos 80) or c1 = (10^7) cos 80

At t = 0, y = 0:

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

And: y’_o = v_o sin (80) = (10^5) sin 80 =

-0.01c4exp(-0.01t) - 100 g

So: c4 = -(10^7) sin 80 – 10000g

therefore, c3 = - c4 exp(-0.01t) = 10^7 sin 80 + 10000g

Now, to obtain the x and y-coordinates:

First, the x-coordinate:

x = c1 + c2 exp(-0.01t) = 10^7 cos 80 –

(10^7 cos 80)exp( -0.01t)

x = 10^7 cos 80 [1 - exp(-0.01t)] =

10^7 (0.17365) [1 - exp(-0.01t)]

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

The y-coordinate:

y = c3 + c4 exp(-0.01t)

y = 10^7 sin 80 + 10000g – 10^7 sin 80

+ 10000g(exp(-0.01t))

y = 10^7 sin 80 + 10000g[1 - exp( -0.01t)]

– 100 gt

y = [10^7 (0.9848) + 320,000][1 - exp(-0.01t)] – 100 gt

y = 1016800(1 - exp(-0.01t) – 100 gt

After 10 seconds (t = 10):

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

x = 165, 200 ft. (= 31. 3 miles)

y = 1016800[1 - exp (-0.01(10)] – 100(32)(10)

y = 930,580 ft. (= 177 miles)

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