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


by tatertot560
Tags: altitude, determine, rocket
tatertot560
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#1
Jan28-09, 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 :)
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LowlyPion
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#2
Jan28-09, 11:09 PM
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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.
Nabeshin
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#3
Jan28-09, 11:11 PM
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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).

tatertot560
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#4
Jan28-09, 11:15 PM
P: 11

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


Quote Quote by LowlyPion View Post
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.
then i use d=vi*t+1/2at^2 right? when i used that equation, i got 38709 m as my answer. is that correct or do i need to rethink what equation i need to use?
LowlyPion
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#5
Jan28-09, 11:20 PM
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I would use the V2 = 2*a*x since initial velocity was at rest.
tatertot560
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#6
Jan28-09, 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?
LowlyPion
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#7
Jan28-09, 11:25 PM
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Quote Quote by Nabeshin View Post
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).
Yes. they have apparently added more info than needed. But the 33 m/s2 apparently governs the issues of weight depletion. The final answer isn't so large as to get into gravitational potentials I should think.

I think the simple path is indicated from the statement of the problem.
LowlyPion
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#8
Jan28-09, 11:27 PM
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Quote Quote by tatertot560 View Post
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?
That's your intermediate answer. (But in m, not km)

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.
tatertot560
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#9
Jan28-09, 11:36 PM
P: 11
additional height? im kind of confused
LowlyPion
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#10
Jan28-09, 11:41 PM
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Quote Quote by tatertot560 View Post
additional height? im kind of confused
By the statement of the problem it goes up 19074 m and it is still going at 1122 m/s. Don't you think it's going to be going higher?
tatertot560
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#11
Jan28-09, 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)
LowlyPion
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#12
Jan28-09, 11:47 PM
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Quote Quote by tatertot560 View Post
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)
No that is incorrect. You apparently added the velocity to the height? They aren't the same thing. Going at that velocity how much more will it go up against gravity?

Keep everything in meters until the end - and then divide by 1000 m to get an answer in km.
tatertot560
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#13
Jan28-09, 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..?
LowlyPion
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#14
Jan28-09, 11:51 PM
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Velocity and height. They aren't the same thing.

Going at that velocity how much more will it go up against gravity?
LowlyPion
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#15
Jan28-09, 11:53 PM
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This time try the same equation

V2 = 2*a*x

Only this time a is gravity slowing you to 0, so a = 9.8
tatertot560
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#16
Jan28-09, 11:53 PM
P: 11
i don't know :(

(thanks for all your help btw)


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