The velocity and height of a rocket launched vertically from the earth.

[Mastablade]

Homework Statement

So if a rocket is launched vertically from the surface of the earth, the rocket has a mass of 1000 kg and has a fuel load of 12000 kg. The fuel burns at such a rate that it will be exhausted after 180 secs. The exhaust velocity of the burned fuel relative to the rocket is 2000 m/sec. Assume an air resistance proportional to the velocity with proportionality constant b = 100 n-sec/m and that the gravity is constant. Find the velocity and height of the rocket when the fuel runs out.

Homework Equations

Force = mass * acceleration = mass * gravity - air resistance * velocity\
Mass = 13000 (weight of rocket and fuel) - 66.66t (weight of fuel lost per second or 12000/180)

The Attempt at a Solution

So, naturally, I want to find the velocity with the above equation. So with inserting the known values into the equation it becomes,
(13,000 - 66.66t)a = (13,000 - 66.66t) * 9.8 (gravitational constant - 100 (air resistance) * v
dividing mass to the other side I get.
a = 9.8 - (100 * v) / (13,000 - 66.66t)
Since I will have to integrate a in get it in terms of v, I divided the right half to the left half
dv/dt (acceleration) / 9.8 - (100 * v) / (13,000 - 66.66t) = 1 dt
Integrate both sides (I had to use a program to solve this)
(-6666t - 130)ln(-653.268t - 100v + 127,400) = t
now we have to get v by itself so,
ln (-653.268t - 100v + 127,400) = t / (-6666t - 130)
Get rid of the ln,
-653.269t - 100v + 127,400 = e ^ (t / (-6666t - 130)
more simple rearranging gets
-100v = e ^ (t / (-6666t - 130) + 653.269t - 127,400
v = -(e ^ (t / (-6666t - 130) + 653.269t - 127,400) / 100
now we have to subtract the exhaust velocity from all this equaling

v = (-(e ^ (t / (-6666t - 130) + 653.269t - 127,400) / 100) - 2000

This should give the velocity of the rocket when the fuel runs out. I theorize that to find the height you simply have to take the integral of this function, and since I don't want to re-write that mess (at least not without knowing if I am actually correct in that assumption), I would greatly appreciate if someone could analyze this work and see if I both found the correct velocity and am correct in how to find the height. Thanks in advance for your help.

 

[AvroArrow]

A:You are almost there. You have a differential equation which you can solve by separation of variables:$$\frac{\mathrm{d}v}{dt} = \frac{\mathrm{d}\left(-\frac{\mathrm{e}^{\frac{t}{-6666t-13000}} - 653.269t + 127400}{100}\right)}{\mathrm{d}t} = \frac{-\frac{1}{-6666t-13000}\mathrm{e}^{\frac{t}{-6666t-13000}} - 653.269}{100} - 2000$$Integrating both sides and using the fact that $v(t=0) = 0$ gives you the following expression for the velocity:$$v(t) = -\frac{1}{100}\left(-\frac{1}{-6666t-13000}\mathrm{e}^{\frac{t}{-6666t-13000}} - 653.269t + 127400\right) - 2000$$Now, to get the height, simply integrate this expression with respect to time.Since you already integrated in the previous step, you can use the fundamental theorem of calculus to write$$h(t) = h(0) + \int_0^t v(\tau)\,\mathrm{d}\tau$$where $h(0)$ is the initial height (which is 0 in our case).The integral is very complicated, but you can easily compute it by using a computer algebra system like Mathematica or Maple.