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Flowerpot going up and down past a window

  1. Aug 22, 2008 #1
    This is a problem from Halliday, Resnick, and Walker. My solution gets the correct answer, but it seems very extremely complicated and I think there is an easier way to do it.

    1. The problem statement, all variables and given/known data

    A cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?

    2. Relevant equations

    I'm not sure, maybe [tex]y(t)=\frac{1}{2}at^2+v_0t+y_0[/tex]

    3. The attempt at a solution

    I assumed that the time the flowerpot was in view going up and the time it was in view going down were equal, therefore they were both 0.25 seconds. How can I justify this?

    Let y(t) be the height of the flowerpot at time t. Let y=0 m represent the top of the window, so y=-2 m would be the bottom of the window.

    Let t=0 be the time the flowerpot first appears in the window, so y(0) = -2. The flowerpot reaches the top of the window at t=0.25 (by the assumption), so y(0.25) = 0.

    Let tmax be the time when the flowerpot is at its highest point. Let ymax = y(tmax)

    Because the acceleration is constant, I know that the equation of the path of the flowerpot, y(t), is a parabola with the vertex at (tmax, ymax).

    Therefore [tex]y(t) = c(t-t_{max})^2+y_{max}[/tex] for some constant c.

    Multiply out y(t) to get: [tex]y(t) = c(t-t_{max})^2+y_{max} = c[t^2-2t_{max}t+t_{max}^2]+y_{max}=(c)t^2+(-2ct_{max})t+(ct_{max}^2+y_{max})[/tex]

    Since the acceleration is due to gravity, the coefficient of t2 must be -4.9, so c=-4.9.

    [tex]y(t) =(-4.9)t^2+(-2 \times -4.9 t_{max})t+(-4.9t_{max}^2+y_{max})=-4.9t^2+(9.8t_{max})t+(-4.9t_{max}^2+y_{max})[/tex]

    We know the values of y(0) and y(0.25):

    [tex]y(0) = -2 = -4.9(0)^2+(9.8t_{max})(0)+(-4.9t_{max}^2+y_{max})=-4.9t_{max}^2+y_{max}[/tex]

    [tex]y(0.25) = 0 = -4.9(0.25)^2+(9.8t_{max})(0.25)+(-4.9t_{max}^2+y_{max})= -4.9(0.25)^2+(9.8t_{max})(0.25)+(-2)=-2.306+2.45t_{max}[/tex]

    From [tex]-2.306+2.45t_{max}=0[/tex], we get [tex]t_{max}= 0.941[/tex] s

    From [tex]-4.9t_{max}^2+y_{max} = -2[/tex], and substituting in tmax, we get [tex]y_{max}=2.34[/tex] meters above the window, which is the correct answer.

    Wow, that took a long time to type and it looks like a mess. Does anyone know a cleaner solution? And how do I conclude that the time going up and down the window are equal? Thanks!
    Last edited: Aug 22, 2008
  2. jcsd
  3. Aug 22, 2008 #2


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    Homework Helper

    First of all it is a uniformly accelerated field. Symmetry should be a good enough argument as per your observation and assumption.

    Secondly I would simplify things by solving it from what you know. The window is 2M. The segment of its flight that kitty can see then is 2M and a .25 s duration. This x and the t, then yields an initial velocity across the segment that kitty can see. (I'm using the case where it's dropping.)

    Knowing initial velocity, then by V = at you know how much time it took to drop from the maximum height and achieve this initial v. This time then when applied once again to x = 1/2 at^2 should yield the distance above the window that the maximum height was.
    Last edited: Aug 22, 2008
  4. Aug 23, 2008 #3
    I am sorry, but I do not quite understand. Do you mean the symmetry of the graph of the position function?

    I think I understand your method, but just to be sure:

    Let v = velocity of flowerpot as it reaches the top of the window (while dropping)

    [tex]\Delta y = -4.9t^2 + vt[/tex]

    [tex]v = \frac{\Delta y +4.9t^2}{t} = \frac{(-2)+4.9(0.25)^2}{0.25}=-6.775 [/tex] m/s

    Let vmax = velocity of flowerpot at maximum height = 0 m/s.

    The time it took the velocity to change from 0 to -6.775 m/s can be found by:

    [tex]v = at+v_{max}[/tex]


    Therefore, falling from velocity of 0 to velocity to -6.775 (from maximum height to level with top of window), the displacement was:

    [tex]\Delta y = -4.9t^2 = -2.34[/tex] m, so the maximum height was 2.34 meters above the window.

    This solution is a lot nicer. Thanks for your help!
  5. Aug 24, 2008 #4


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    There's another way to do this by applying the symmetry of the flowerpot's trajectory. You know it came to a stop at some height H above the top of the window, which is the distance you are asked to find. The pot takes some unknown time T to do fall back down that far. Thus, falling from rest, and calling downward positive and the top of the trajectory y = 0, we have

    [tex]H = \frac{1}{2} \cdot gT^2[/tex]

    As has also been argued by symmetry, the pot drops out of sight past the bottom of the window a quarter-second later, at T + (1/4) seconds. At that moment, it is 2 meters lower, at y = H + 2 meters, so we can write

    [tex]H + 2 = \frac{1}{2} \cdot g \cdot (T + \frac{1}{4})^2

    = \frac{1}{2} \cdot g \cdot (T^2 + \frac{1}{2} \cdot T + \frac{1}{16})[/tex]

    We now substitute H from the first equation to solve for T:

    [tex]\frac{1}{2} \cdot gT^2 + 2 = \frac{1}{2} \cdot g \cdot (T^2 + \frac{1}{2} \cdot T + \frac{1}{16})[/tex]

    [tex]\Rightarrow T = \frac{8}{g} - \frac{1}{8} \approx 0.690 seconds[/tex]

    This is how long the flowerpot was falling from rest at the top of its trajectory down to the top of the window, so that distance is

    [tex]H = \frac{1}{2} \cdot g \cdot (0.690^2) \approx 2.34 meters[/tex]
    Last edited: Aug 24, 2008
  6. Aug 24, 2008 #5


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    No my reference was to the fact that your pot was in motion in a uniformly accelerated field - gravity.

    You know the saying that what goes up must go down. And you know generally that there must be a conservation of energy. Hence whatever kinetic energy the pot possessed on it's upward flight at a particular level, it will possess in its downward at the same level. If that is true at one arbitrary point, say the level of the bottom of the windowsill, then by induction it should also be true at every level up - all the way to the top of its trajectory - as well as below failing any perturbation. Therefore if the speeds match - save for signs - up and down, then time to traverse is likewise constrained to be the same. This also constrains your position function to be symmetrical too.

    Hence you can exploit symmetry to break the problem into more manageable - read that as easier - pieces.
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