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Difficult Problem I Tried My Best

  1. Aug 26, 2007 #1
    Difficult Problem!!! I Tried My Best!!

    1. The problem statement, all variables and given/known data
    During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.50 m/s^2. When it is 230 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance is ignored)

    a. How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change it acceleration?

    b. What total distance did the canister travel between its release and its crash onto the launch pad?

    2. Relevant equations
    I know that I have to use one of the constant accelertion, but which one?

    Vx = Vox + axt
    x = Xo + Voxt + (1/2)ax(t^2)
    Vx^2 = Vox^2 + 2ax(x-xo)
    x-xo = ((Vox + Vx)/(2))*t

    3. The attempt at a solution

    Thanks for any help provided.
  2. jcsd
  3. Aug 26, 2007 #2


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    Well, you can start by writing down every piece of information you have gathered from the story problem.

    A good way to organize the information is to write down every relevant events (such as "rocket takes off" or "canister hits ground"), and then organize the information that way -- e.g. write down the velocity of an object at a particular event, or the time of an event, or the acceleration it underwent between events, et cetera.

    It might also be useful to include on this chart the things you want to find (and name them with variables), or you simply think might be useful to understanding the setting or solving the problem.
    Last edited: Aug 26, 2007
  4. Aug 26, 2007 #3
    First you might like to find how long it takes the canister to hit the ground.

    To do that, think of which equation which relates to vertical motion you need to use.

    We also need to know 'u' which is the initial velocity when it starts to fall down.

    Since we don't know time, we have to find the final velocity. And from that work your way through the equations which you stated above to find the time and other unknowns u have to find, you have all the equations, it's really not more than 4 - 5 lines. Think at each part which unknown you need to find out and which equation is best suited

    So once we know how long it takes the canister to fall down to the launch pad. We must think in that given time, how high has the rocket now risen?

    Last edited: Aug 26, 2007
  5. Aug 26, 2007 #4
    I'd approach this by finding the time for the cannister to reach the ground, then applying this time to the rocket.

    For part B, use what you've done in part A.

    Hint: the cannister will NOT drop as soon as it is released, as it still has momentum.

    Edit: there Hurky. :)
    Last edited: Aug 26, 2007
  6. Aug 26, 2007 #5


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    Of course, you're just giving him answers. :grumpy: He needs to learn how to look at the information he's given and the information he wants, and from that discover a plan for solving a problem! Just telling him the plan doesn't help him with that.
  7. Aug 26, 2007 #6
    Sorry, i'll edit mine out
  8. Aug 27, 2007 #7
    Thanks for the answers and I am not a "Him", LOl. Ok but would this work:
    ok I know that the rocket is going to go 230 m when the canister will be released.
    I used the equation:
    230 = (-1/2)(-9.8)(t^2)
    t = 7.06 s. Then I can apply the equation change in x = Voxt + 1/2at^2.
    that would be:
    change in x = (1/2)(3.5)(7.06^2)
    I get 87.2 and I would add the 230 to it and that would be 317.2. but this is not the answer because the software denies it. so where do you think I am going wrong?
  9. Aug 27, 2007 #8
    now, going to try this one ^^

    230 = (-1/2)(-9.8)(t^2)

    but didn't canister have a velocity when it was released?

    x = (1/2)(3.5)(7.06^2)

    and I think you are assuming that v0x is 0,
    how you know that the rocket has been acceleration for 230 m, so i must have a v0>0..

    hopefully, I am right.
    Last edited: Aug 27, 2007
  10. Aug 27, 2007 #9
    thanks in advance for trying..
  11. Aug 27, 2007 #10
    ok, so when you solve for the x = you get 82.1. ANd then I add that to 230 to get the height when the canister will actually hit the ground, so, the answer then is 312.1 but that is no where near the answer. What should i do??
  12. Aug 27, 2007 #11
    oh yeah, the velocity is not of the canister but of the rocket...
  13. Aug 27, 2007 #12
    but canister is a part of rocket, so when the rocket discards it, it has a v of >0. (**.12),
    and t>10s.. I guess so
    Last edited: Aug 27, 2007
  14. Aug 28, 2007 #13
    Good man, you're thinking. This is how to become good with physics.
  15. Aug 28, 2007 #14
    The time it takes for the rocket to reach 230 m (use x = 1/2at^2, v0x = 0, and x0 = 0 because the rocket started from rest on the launch pad):

    230 = 1/2 (3.50) t^2 ==> t=115 seconds

    The time it takes the canister to hit the ground:

    230 = -1/2 (-9.8) t^2 ==> t= 6.9 seconds

    My reasoning is this: You know that when the rocket drops the canister, the height of the rocket is 230 m. Since it takes 6.9 seconds for the canister to hit the ground, how much more will the rocket have traveled in this time?
    Last edited: Aug 28, 2007
  16. Aug 28, 2007 #15
    mrlucky: You haven't included the momentum the cannister has when it's released, this means it still has a velocity equal to the velocity of the shuttle at the instant when it was released. The rocket doesn't 'drop' the cannister.
  17. Aug 28, 2007 #16
    Yeah I thought about it for a momment and you're right. The momment the canister gets dropped, it's velocity is actually positive even thought its acceleration is negative. I'm curious to see the solution myself. Thanks for pointing that out.
  18. Aug 28, 2007 #17


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    Get the velocity at 230m... and everything else will fall into place.
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