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Kinematics and Vectors

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 3.70 (m/s^2}. At 7.0 (s} after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.

    a)How high above the launch pad will the rocket eventually go?
    b)Find the rocket's velocity at its highest point.
    c)Find the magnitude of the rocket's acceleration at its highest point.
    d)Find the direction of the rocket's acceleration at its highest point.
    e)How long after it was launched will the rocket fall back to the launch pad?
    f)How fast will it be moving when it does so?

    2. Relevant equations

    a)d=vot+1/2at^2
    b)v=vo+at
    c)c^2=a^2+b^2; magnitude
    d)no equation needed (?)
    e)t=vf+vo/a
    f)Unsure of what is used here


    3. The attempt at a solution

    Well, I'm unsure of my chosen equations. Please, bear with me.

    d=0(7.0s)+1/2(3.70m/s^2)*(7.0s^2)
    v=0m/s+(3.70m/s^2)(7.0s)
    Cannot seem to make sense of x and y to solve for magnitude

    Looking for a general guidance and explanation of things.. not just an answer. Thanks in advance!
     
  2. jcsd
  3. Sep 13, 2007 #2
    Please help...
     
  4. Sep 13, 2007 #3
    Okay... so I'm not sure where exactly to start with all of this. It seems to be a big problem with manipulating these equations for me. Can someone please make sense of this problem... I don't want the answer!!
     
  5. Sep 13, 2007 #4
    For part b, highest point should be when velocity drops to 0 due to gravity, hence v = 0. unless they are asking for maximum velocity.
    If so, for part c, acceleration will be due to gravity, as there is no upward thrust
    part e, find the maximum height first, that would be the distance the rocket needs to travel with the help of gravity.
    part f, all will be clear when you solve for part e
     
  6. Sep 13, 2007 #5
    When trying to solve a) .. I arrived at an incorrect answer using d=0(7.0s)+1/2(3.70m/s^2)*(7.0s^2). Not sure what else I could possibly use. I understand what is meant by no upward thrust since it fails. So, 9.8 m/s^2 was correct. The direction is definitely downward because of this. Still not clear about part e.
     
  7. Sep 13, 2007 #6
    Okay, so I feel as if I'm getting the cold shoulder here. I honestly do not understand regardless of answers which is why I indicated I wasn't intending on someone doing the work for me... Anyone else care to explain the logic in all of this?
     
  8. Sep 13, 2007 #7
    A simple way for me to understand is to sketch the velocity time graph. Area under the graph will give you the displacement. All the formulas can be derived from the graph and it should help you in understanding, it helped me when I started learning too. And it saves you a lot of time thinking which formula to use.

    For part a perhaps you need to include the remaining distance when the velocity decreased from its peak to 0, as it is still moving upwards.

    Using the graph, part e will need you to extend the graph to below the time axis and find y where the area from t=0 to the t=x (where x is the time when velocity is 0) is the same as the area from t=x to t=y, as this would mean the positive displacement = negative displacement. Part f will be finding that negative velocity.
     
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