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Free Fall of a flower pot

  1. Feb 22, 2009 #1
    Having trouble with the equation to solve the problem.

    A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally in possession of a high-precision timing system, notices that it takes 0.20 seconds for the pot to fall past his 4.0 m high window. How far above he top of the window is the ledge from which the pot fell? ( Neglect any effects due to air resistance.)
  2. jcsd
  3. Feb 22, 2009 #2


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    What equation relates distance time and acceleration directly?
  4. Feb 22, 2009 #3
    d= v1(t)+1/2(a)(t) squared

    d= ((v1+v2)/2) * t

    is that correct?
  5. Feb 22, 2009 #4
    Where you get v2 from?

    There is only 1 thing moving. That second equation is some kind of average. Most texts have the free fall equations in them already derived. The first one you have is what you need. Now just think about what the velocity means and you can get your distance.

  6. Feb 22, 2009 #5


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    The first one is useful. So let's use it.

    Can't you determine what the height is it fell from rest to the top of the window?

    d = 1/2*g*t2

    and then can't you also write another equation for when it passes the bottom of the window?

    d + 4 = 1/2*g*(t + .2)2

    Glory be. 2 equations and 2 unknowns. The answer can't be far away now.
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