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Constant acceleration of velocity

  1. Sep 28, 2005 #1
    A falling stone takes 0.28s to travel past a window 2.2m tall. From what height above the top of the window did the stone fall?

    So we have time and dinstance, but not the acceleration, which confuses me terribly. Do I need to find velocity? Or do I need to look for the acceleration? SO confused. Any help will be greatly appreciated.

    Thanks!!
     
  2. jcsd
  3. Sep 28, 2005 #2

    hotvette

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    It always helps to draw a picture and free body diagram, then write the equation of motion, and solve for what you want. In this case, you have a rock of mass m, acted upon by gravity g, falling from rest, starting from a height x above the window. What you are given is the time it takes to cover a certain distance during it's fall to the ground. Hope this helps get you thinking in the right direction.
     
  4. Sep 28, 2005 #3
    I appreciate your advice, but I am still confused. I am using the equation for motion, yes. But it asks for time, and the time from the moment of fall is not the same as the time it took the rock to fall along the window, which means that t is the variable that cannot be used, but is needed for the equation to find the distance covered. And, yes, drawing a diagram is the first thing I do... :)
     
  5. Sep 28, 2005 #4

    hotvette

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    Let me try this way. You have an equation of motion that relates distance fallen (from an unknown point above the window) as a function of time. Seems to me a fruitful approach is to express that equation of motion for [itex]x_1, t_1[/itex] and [itex]x_2, t_2[/itex], where [itex]x_1[/itex] represents the top of the window and [itex]x_2[/itex] represents the bottom of the window. Subtracting the 2 equations gives you [itex]x_1 - x_2[/itex], which you know.
     
    Last edited: Sep 29, 2005
  6. Sep 29, 2005 #5
    Ok, what if we don't even use time, except to find the velocity? Having that information, we can find the position, assuming that the acceleration is -9.8m/s^2?
     
  7. Sep 29, 2005 #6

    hotvette

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    Had to do a little head scratching. I believe the following will work:

    Using the equation of motion of the rock you should be able to get an equation for [itex]x_1-x_2[/itex] in terms of [itex]g[/itex], [itex]t_1^2[/itex] and [itex]t_2^2[/itex]. You know [itex]x_1-x_2[/itex], and [itex]g[/itex], but not [itex]t_1[/itex] or [itex]t_2[/itex]. But, you do know [itex]t1-t2[/itex]. Thus, you have two equations in two unkowns and should be able to solve for [itex]t_1[/itex], back substitute into the equation of motion, and find [itex]x_1[/itex].
     
    Last edited: Sep 29, 2005
  8. Sep 30, 2005 #7

    hotvette

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    Were you able to solve it? I got [itex]x_1[/itex] equal to a little over 2m using the above methodology.
     
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