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Vertical Motion

  1. Mar 10, 2005 #1
    "With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (aprox. 550 feet)?" Given acceleration in regards to time a(t) = -32 feet per second due to gravity. Neglect air restistance.

    What I've thought of, done etc etc
    a(t)=-32 (which is f'')
    f=-16t^2+(Initial Velocity)t+0

    I know I need to use the 550ft somewhere, either as the f, or maybe set velocity (anti-derivative of acceleration, v=-32t+c, which c is what I'm looking for) I think I could solve if I had t, time in seconds. But not a clue how. Things I tried were trying to solve for t using the position equation (I use the f function for position) But had no luck (much erasing, and the backside of a paper later) solving for when velocity=0 (0=-32t+c) yet thats what I'm looking for.

    Any help would be greatly appreciated, a 'push' in the right direction would make my day =)

    Thanks ahead, DV
     
  2. jcsd
  3. Mar 10, 2005 #2

    arildno

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    Well, in your version, you need the two equations:
    [tex]v(t)=v_{0}+at[/tex]
    [tex]h(t)=v_{0}t+\frac{a}{2}t^{2}[/tex]
    You are to solve these equations for [tex]v_{0}, t*[/tex]
    so that v(t*)=0, h(t*)=H, where H is the height of the building.

    Alternatively, you could determine [tex]v_{0}[/tex] directly from conservation of mechanical energy, without having to determine the actual instant when it occurs.
     
  4. Mar 10, 2005 #3
    Awesome! Thanks so much.
    550=(0)+32/2t^2
    34.375=t^2
    5.863=t

    0=-32(5.863)+C
    0=-187.616+C
    187.616=C

    Which is the answer in the book! Thanks again arildno.
     
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