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Rocket Question

Hi !

I seem to get the right answer when I round up, and use g=10 instead of 9.81... but I don't think I am quite correct.
I have shown it doesn't lift off immediately since the weight is greater than the upthrust initially.
I know the time taken should be for the weight to balance the upthrust exactly, I found the difference in weight, divided by g and then divided by the 16 rate. but this would take into account the fuel AND rocket? The units work out and kind of so does the answer i got of 1.1 (rounded) but I'm a little confused. could someone please provide a solution to this?
Many thanks :smile:
IMG_9346.JPG
Original post by marinacalder
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Can we get some more dimensions on that pic...?
Original post by RDKGames
Can we get some more dimensions on that pic...?


IMG_2023.jpg
Original post by marinacalder
Hi !

I seem to get the right answer when I round up, and use g=10 instead of 9.81... but I don't think I am quite correct.
I have shown it doesn't lift off immediately since the weight is greater than the upthrust initially.
I know the time taken should be for the weight to balance the upthrust exactly, I found the difference in weight, divided by g and then divided by the 16 rate. but this would take into account the fuel AND rocket?


I'm not sure precisely what you're asking. Can you put up your working, please? You need to take into account the mass of the rocket+fuel, but it's only the mass of the fuel that changes, and you know dm/dt for that, so I can't see where the problem would lie.
so its ok to use that rate along with the total mass of rocket and fuel? since the rocket mass doesn't change?
(edited 6 years ago)
Original post by marinacalder
so its ok to use that rate along with the total mass of rocket and fuel? since the rocket mass doesn't change?


The rate of change of mass is dMRdt+dMFdt=dMFdt\frac{dM_R}{dt}+\frac{dM_F}{dt}=\frac{dM_F}{dt} since the mass of the rocket doesn't change. So yes, it's fine to use the given rate of change on its own.
Original post by atsruser
The rate of change of mass is dMRdt+dMFdt=dMFdt\frac{dM_R}{dt}+\frac{dM_F}{dt}=\frac{dM_F}{dt} since the mass of the rocket doesn't change. So yes, it's fine to use the given rate of change on its own.

YAY! many thanks :smile:

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