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Verifying why the equation x=x0+v0t+(1/2)at^2 produces 2 times when solving for t

  1. Aug 10, 2012 #1
    Could anyone please verify with me that I have the right idea in answering the question below.

    1. The problem statement, all variables and given/known data
    If you know the initial velocity v0 and the initial and final heights y0 and y, you can use x=x0+v0t+(1/2)at^2 to solve for the time t when the object will be at height y. But the equation is quadratic in t, so you'll get two answers. Physically, why is this?

    3. The attempt at a solution
    This occurs because during the motion of an object it has passed through it's initial displacement on 2 occassions. Thus times occur twice.
     
  2. jcsd
  3. Aug 10, 2012 #2

    CWatters

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    One on the way up and the other on the way down probably.
     
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