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Integration and initial velocity

  1. Jan 26, 2009 #1
    1. The problem statement, all variables and given/known data

    With what initial velocity must an object be thrown upward (from ground level) to reach a maximum height of 550feet.

    Use a(t)= -32ft/sec2 as the acceleration due to gravity. (neglect air resistance)

    2. Relevant equations

    Use integration

    3. The attempt at a solution

    I know that U should first start off by integrating the acceleration in order to get velocity, but I wind up getting:

    32x+C= v(t)

    I'm not sure how to deal wit the problem from here, does anyone have any suggestions?
     
  2. jcsd
  3. Jan 26, 2009 #2

    CompuChip

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    What is x?
     
  4. Jan 26, 2009 #3
    Im pretty sure that x is suppose to be time, because that is the only thing that relates acceleration and velocity
     
  5. Jan 27, 2009 #4

    CompuChip

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    I thought t was time as well. So lesson one is to be a bit more precise in your notation.

    a(t) = -32 ft/s2, then
    [tex]v(t) = \int a(t) \, dt[/tex]
    where t is the variable. Integrating a constant over t gives you the constant times t so
    v(t)[ft/s] = - 32 t + C

    Note the minus sign, which is carried over from a(t) < 0.
    C is an integration constant which you need to determine. What condition will you use for this?
    How can you see in the v(t) graph or formula that the highest point is reached?
     
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