# Vertically launched rocket problem with integration

1. Oct 3, 2011

### Lawrencel2

1. The problem statement, all variables and given/known data
Find the Rockets height as a function of time
It is in a constant field g. u refers to the exhaust speed and M is the initial mass.
Starts from rest. and is single stage.
2. Relevant equations
1) m * dv/dt= -dm/dt * u-mg

2)Show that height as t is: y(t)= u*t- 1/2*g*t^2- u*t*ln(M/m)

3. The attempt at a solution
Ok, i arrived at a function very similar to the the height but, i cannot seem to get the u*t term in the function. i get y(t)= - 1/2*g*t^2- u*t*ln(M/m)
where do i get the u*t term from? i feel like i am so close to the answer but so far away..
I integrated 1) and arrived at V(t)= u*ln(M/m)-g*t

2. Oct 4, 2011

no help?

3. Oct 4, 2011

### gash789

Hi, I did look at this earlier but got stuck with your first integration. It is likely there is a constant of integration you are missing so if you are still stuck and want to, please can you show me your steps to integrate 1) and then I will see if I can help?

Cheers

4. Oct 6, 2011

### PICsmith

I'm pretty sure you're forgetting that mass is a function of time, so your integration of u*ln(M/m(t))*dt isn't as simple as you had hoped...

Edit:
Hint: Do the integration w.r.t. mass, and not time. The key here is that (I'm assuming) the fuel burn rate is constant, so dm/dt = -c, where c is some constant. Use that to replace dt with -dm/c. Also, don't forget to solve for the constant of integration using the initial condition m(0) = M.

Last edited: Oct 6, 2011