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Impulse and maximum height of a particle

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle receives an impulse that lasts 1s, coming from a upwards vertical force. This force is given by the following equation: F = -8t²+8t.

    What is the maximum height reached by the particle?


    2. Relevant equations

    F = -8t²+8t

    3. The attempt at a solution

    [itex]\int(-8t²+8t)dt[/itex] = [itex](-8t³ +12t²)/3[/itex]

    When t = 1s, I = 4/3. That's what I got as my impulse.

    For the maximum height, I took the derivative of F and set it = 0, finding that t = 1/2

    Now I don't know how to find the height with all the information I have.
     
  2. jcsd
  3. Feb 16, 2014 #2

    SammyS

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    Hello thaalescosta. Welcome to PF !

    (Use ^2 for an eponent of 2 in Latex.)

    There's no need to use the derivative of the force, F.

    The change in momentum of the particle is equal to the impulse. (Impulse - Momentum Theorem)


    Do you know the mass of the particle?
     
  4. Feb 16, 2014 #3
    Got it :)

    So I = ΔP = P[itex]_{f}[/itex] - P[itex]_{i}[/itex] = ∫Fdt, from i to f

    My impulse turned out to be 4/3.

    The mass of the particle wasn't given. I'm guessing that the answer will be in general terms.

    If I = m.v, then v = 3m/4

    And if the maximum height is
    h[itex]_{max}[/itex] = v²/2g
    then
    h[itex]_{max}[/itex] = 9m²/32g


    Does this make sense?
     
  5. Feb 16, 2014 #4

    SammyS

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    Yes.

    You probably should enclose the entire denominator in parentheses.

    [itex]h_{max} = v^2/(2g)[/itex]

    [itex]h_{max} = 9m^2/(32g)[/itex]
     
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