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Free fall 2

  1. Feb 2, 2008 #1
    1. A flower pot falls out of a window and past the window below (ignore air resistance). It takes the post .42 s to pass this window, which is 1.9m high. How far is the top of the window below the windowsill from which the flowerpot fell?
    answer: .31 m




    2.
    v = v0 + at
    x = x0 + v0t + 1/2at^2
    v^2 = v0^2 + 2a(x - x0)




    3. The attempt at a solution
    Known facts:
    t = .42 s
    bottom window height = 1.8
    a = -g
    x0 = 0 (if the origin is the top windowsill)
    v0 = ?
    x = ?

    x = 0 + v0(.42) + 1/2(-9.8)(.42)^2

    So I need to find the initial velocity so I can get the total distance and subtract it from the 1.9. If I had the total distance down, then I could get it easy, but I'm stumped from here since I'm trying to find the total distance. I've tried using v0=0 and that didn't work. I also tried setting the v = 0 (thinking I could do that sense the problem ends with that bottom window). Every route I go it seems I'm still left with two unknowns or I end with a total distance of .86 which isn't even longer than the bottom window.


    Any hints on which equations to put together?
     
  2. jcsd
  3. Feb 2, 2008 #2
    hi there:

    By conservation of energy:
    init=final
    mgh=1/2mv^2
    v^2=2gh
    That means that he reach the top of the window with v velocity.
    So, we can rewrite the equation of motion as:
    (v and g minus because the referential)
    x=xi-sqrt(2gh)t-1/2gt^2
    x-xi is given:
    x is the bottom of the window
    xi is the top of the window
    t-ti is given(consider ti=0 in top of window)

    h is computed: 1 var with 1 equation...

    Right?
     
  4. Feb 2, 2008 #3
    Thanks, but you've went a little over my head. We haven't covered anything but the basic equations. To be honest, it seems like there's something missing in the question. Either the total height needs to be given or the velocity...
     
  5. Feb 2, 2008 #4
    ok, you know by conservation of mechanical energy that:

    inicial Energy = Final Energy

    well, Energy=Kinectic+Potencial

    so, inicial K+inicial P=final K+final P
    but
    [tex]K=0.5*m*v^{2}[/tex]
    [tex]P=m*g*x[/tex]

    you can consider that, in first step, the x=0 in the windowsill, and the inicial velocity=0,and -xf the distance from windowsill to the top of the window below,
    so:
    [tex]m*g*-x_{f}=0.5*m*v_{f}^{2}[/tex]
    compute in order to v and you get the velocity in the top of the window below.

    Now, we know the velocity in the moment the plant reach the window top as function of h(the heigh between the windowstill and the window top.

    so, we make the second step: find the velocity needed in the window top so that the plant need 0.48s to reach the end.

    to do that, you use the equation of movement, as you have the xi, the t, and the g
    so, you can compute v.
    x=xi+vo*t-0.5g*t^2
    t is given(as you consider t=0 in window top), xi is 1.9m, x=0, g is 9.8. vo is computed

    if no, pm me and we get it trough msn or something
     
    Last edited: Feb 2, 2008
  6. Feb 2, 2008 #5
    my professor solved this problem the other day with the 3 formulas you stated, very above average difficulty. very
    (for intro physics, its overwhelming), maybe littlepigs method is way better, but that takes a little more indepth

    ps, he used this problem to introduce the possibility that you can arrange your problem so that T = T - delta T
    (delta T being the time that it takes to cross the window, T being the total travel time)
     
    Last edited: Feb 2, 2008
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