
#1
Jan2809, 11:01 PM

P: 11

1. The problem statement, all variables and given/known data
A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 33 m/s^2 for 34 s, then runs out of fuel. Ignroe any air resistance effects. A) what is the rocket's maximum altitude? B) how long is the rocket in the air? 2. Relevant equations i tried using vf=vi + at for part a and then i think i use the equation d=vi*t+1/2*a*t^2 to find the altitude. 3. The attempt at a solution for vf, i got 1122 m/s, but i'm not sure if this is right or if i'm using the right equations. if that is correct, i plug it into the 2nd equation to get my second answer, right? if anyone can help, i would really appreciate it :) 



#2
Jan2809, 11:09 PM

HW Helper
P: 5,346

Welcome to PF.
That's correct. V = a*t = 33*34 = 1122 Now how high is it at the point of main engine shutoff? After that it's simply a matter of figuring how much longer it goes up against gravity. And then figuring that distance after shutoff and adding the 2 heights together. 



#3
Jan2809, 11:11 PM

Sci Advisor
P: 2,194

The equations you're using are correct. However, the method of solution you suggest is to describe a situation where the rocket leaves the ground with an initial velocity equal to that after a 34 second burn, and then is under solely the influence of gravity. You have to remember that during the burn time, the rocket is also moving upwards!
P.S: This problem is worded rather strangely. The mention of fuel makes me think acceleration shouldn't be constant (Rather, force would be constant). Additionally, I'm not sure if you're supposed to consider that g is not constant (as it may well make a significant difference for a problem such as this). 



#4
Jan2809, 11:15 PM

P: 11

determine a rocket's altitude; how long rocket is in air 



#5
Jan2809, 11:20 PM

HW Helper
P: 5,346

I would use the V^{2} = 2*a*x since initial velocity was at rest.




#6
Jan2809, 11:23 PM

P: 11

okay. i used that equation
v^2=2*a*t (1122)^2=2(33)x x=19074 km and i got that as my answer. that's correct, right? 



#7
Jan2809, 11:25 PM

HW Helper
P: 5,346

I think the simple path is indicated from the statement of the problem. 



#8
Jan2809, 11:27 PM

HW Helper
P: 5,346

Because now you have yourself a rocket going 1122 m/s at a height of 19km. So now figure the additional height and add the 2 together. 



#9
Jan2809, 11:36 PM

P: 11

additional height? im kind of confused




#10
Jan2809, 11:41 PM

HW Helper
P: 5,346





#11
Jan2809, 11:43 PM

P: 11

so i add those two together right? its asking for the answer in km. so if i add the two distances together and multiply by 1000, i'll get the right answer? (i already have an answer but i'm not sure if it's right. i got 20196000 km)




#12
Jan2809, 11:47 PM

HW Helper
P: 5,346

Keep everything in meters until the end  and then divide by 1000 m to get an answer in km. 



#13
Jan2809, 11:49 PM

P: 11

i added 1122 m/s and 19074 together then multiplied by 1000 to get km...wait, i did that wrong didn't i? do i have to use another equation to figure out the other height..?




#14
Jan2809, 11:51 PM

HW Helper
P: 5,346

Velocity and height. They aren't the same thing.
Going at that velocity how much more will it go up against gravity? 



#15
Jan2809, 11:53 PM

HW Helper
P: 5,346

This time try the same equation
V^{2} = 2*a*x Only this time a is gravity slowing you to 0, so a = 9.8 



#16
Jan2809, 11:53 PM

P: 11

i don't know :(
(thanks for all your help btw) 


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