# Simple Problem. Having trouble

1. Oct 2, 2005

### antiflag403

Hey everyone!
Ok. Im having trouble with a problem here. Please dont post the answer just help me into the right direction.
A 200 Kg weather rocket is loaded with 100 Kg of fuel and fired staight up. it acclerates upward at 30 (m/s)/s for 30 s, then runs out of fuel. Ignore any air resistance effects.
a) what is maximum altitude?
b) how long is the rocket in the air?

Ok. So I'm pretty sure that the masses are put in to throw you off, and not necessary for the problem. I am having trouble because I am not sure what the upward acceleration would be. Do I just use 30 (m/s)/s and neglect gravity while it goes up, then use only gravity for it falling?? If this is true I can just use simple equations to determine the answers.
Thanks for any help!!!!

2. Oct 2, 2005

### antiflag403

Ok. Ive done more thinking and think that I have to find out how high and how fast it gets while accelerating at 30 (m/s)/s. Then with this velocity find out how high it continues up while accelrating in the negative direction (due to gravity). This total displacement will be the answer to a. Is this correct???

3. Oct 2, 2005

### Staff: Mentor

Yes, that's correct.

4. Oct 2, 2005

### antiflag403

Ok. So I did that and got the seemingly ridiculous answer of 54.8 Km. Can someone check to see if I went wrong somewhere??? As well, i am having some trouble determining the total time the rocket is in the air. I think it went upwards for 121.8 s, but am not sure how to figure out how long it takes go get down. Again, any help is greatly appreciated.

Last edited: Oct 2, 2005
5. Oct 2, 2005

### Staff: Mentor

Your answers are correct. (It's a rocket, after all!) To find the time of descent, just treat it as any other falling object.

6. Oct 2, 2005

### antiflag403

THANKS ALOT DOC!!!!
So i did the calculations as i think they should be done and came up with the time of decent being 83.5 seconds, which brings the total time in the air to 205.3 seconds. Hopefully this is right.
As well we had to draw a velocity-time graph to go along with the problem. Am i right in thinking that the graph will slope sharply upwards for 30s, starting at the origin and ending at 900. Then the graph slopes more gently all the way past the origin into the negative area before making a sudden jump back up to zero??????
Again thanks alot to doc for all the help!

7. Oct 3, 2005

### Staff: Mentor

Recheck this time. Remember that it must fall the full distance from its maximum altitude to the ground.

8. Oct 3, 2005

### antiflag403

I re-did the calculations and think i found my mistake. My new time of decent is 74.76s bringing the total to 196.56s. I obtained this by taking:
a=V2-V1/t(square) and solving for t. Can i do this????
Again thanks for the help Doc

9. Oct 3, 2005

### Staff: Mentor

That equation makes no sense to me. (Check the units!) For one thing, $a = \Delta v / \Delta t$, but that won't help you here, since all you know is the distance, not the final speed.

What's the formula for distance as a function of time for a falling body?