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Vertical Motion

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A toy rocket is projected vertically upwards from rest and it rises with an acceleration of (9-5t)g m / s^2 for the first two seconds and, thereafter, freely against gravity. Find: the maximum speed and the maximum height.


    2. Relevant equations
    a = (9-5t)g for the first 2 seconds.
    general motion equations

    3. The attempt at a solution
    The acceleration after 2 seconds is (9 - 10). -g = 9.81 m/s^2.
    The velocity at this point is:
    v = u + at
    v = 0 + 9.81 x 2 = 19.62 m/s.
    Max height occurs when v = 0 therefore ;
    0 = 19.62 -9.81 x t
    t is = 2, which gives a total time of 4 seconds.
    I can get the the max height and velocity easily from here. Except the answers are quite bigger than what I'm getting. So I'm thinking I'm doing something wrong for these first 2 seconds. I don't think I can integrate the acceleration to find velocity because I don't have enough information.

    Any help is appreciated :)

    Noir
     
  2. jcsd
  3. Nov 7, 2009 #2

    rl.bhat

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    Homework Helper

    Up to 2 seconds, the acceleration is varying. So
    dv = ( 9-5t)*g*dt.
    To get the velocity at 2 seconds,integrate dv between 0 to 2 seconds. Then proceed.
     
  4. Nov 7, 2009 #3

    Delphi51

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    Homework Helper

    It looks like you are using the formula Vf = Vi + a*t here.
    But that formula only applies when the acceleration is constant.
    Your acceleration varies with time during the first two seconds.
    You can deal with varying acceleration by using calculus (integration) or by drawing an acceleration vs time graph using your knowledge that the area under the graph is the velocity.
     
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