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Free falling

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A flowerpot falls past two meter tall window in ¼ of a second.

    2. Relevant equations

    From what height did the flowerpot fall?

    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2009 #2

    Pengwuino

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    What is your attempt at the solution?
     
  4. Oct 9, 2009 #3
    I can calculate the velocity of the pot while this is passing the window.
    v = d/t = 2m / (1/4s) = 8 m/s

    But how can this help me now?

    Do I simply plug the velocity into the formula: h = v^2 / 2g?
     
    Last edited: Oct 9, 2009
  5. Oct 9, 2009 #4

    Pengwuino

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    Well remember, that only applies for constant velocities. You can use your kinematic equations to see what initial velocity an object must have to travel 2m in .25 seconds under a constant acceleration caused by gravity.
     
  6. Oct 9, 2009 #5
    So the proper formula to use in this case would be:

    s = ut + at^2 / 2, where u= 8 m/s and t= 1/4 s?
     
    Last edited: Oct 9, 2009
  7. Oct 9, 2009 #6

    Pengwuino

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    No, the velocity is constantly changing so do not use 8m/s. You know the height of the window and the time it took to go that distance. You can then solve for the initial velocity that the object has when it reaches the window. This question is poorly worded so if you assume the object was simply dropped from a distance above the window, you can determine the distance above the window it was dropped using [tex]v^2 = v_0^2 + 2 a \Delta x [/tex] where [tex]v_0[/tex] is 0 since we're looking for the distance covered if it was dropped as opposed to thrown downward.
     
  8. Oct 10, 2009 #7
    Since the initial velocity is 0, do we plug the 8m/s in place of the v squared, so the equation would be (v)squared= ( 0 ) squared + 2(9.8)x and the final answer should be 3.27m, right?!?!
     
    Last edited: Oct 11, 2009
  9. Oct 11, 2009 #8

    Pengwuino

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    No, don't use 8m/s anywhere in this problem. You know that:

    [tex]\Delta x = v_0t + \frac{at^2}{2}[/tex]. You know the window is 2m, which is your [tex]\Delta x[/tex]. You know the time and the acceleration, so determine the initial velocity the object has when it reaches the top of the window.

    With this in mind, you can then use [tex]v^2 = v_0^2 + 2 a \Delta x [/tex] knowing [tex]v_0[/tex] is 0 at the moment the object is dropped and you know that the final velocity will be your initial velocity from the previous section. With this, you calculate the [tex]\Delta x[/tex] which is simply the length above the window the object was dropped. Add that to the length of the window and you have your answer.
     
  10. Oct 11, 2009 #9
    OK, so after plugging the numbers into the given formula, I got the result for
    v0= 6.67 m/s.
    And putting the v0 for v into the second formula gave me the result x= 2.03 m and adding to this the length of the window, I get the final result which is 4.03 m.
    Did I calculate correct this time?
     
  11. Oct 11, 2009 #10

    Pengwuino

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    The velocity when it reaches the window is 6.7m/s, yes. However, I calculated the height above the window being 2.3, not 2.03. Recheck your calculation, sounds like you've about got it.
     
  12. Oct 11, 2009 #11
    Thank for your help! It's all around us and it happens every day, but it is still so confusing when it's put into a real physical problem with numbers and formulas!
     
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