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A dropped stone falls past a window, determine it's initial height

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A falling stone takes 0.33 s to travel past a window 2.2 m tall.
    upload_2014-9-30_14-18-16.png
    From what height above the top of the window did the stone fall?
    Express your answer using two significant figures.

    2. Relevant equations
    Kinematic Equations ~
    X= Vi (time) + .5gt^2

    Vf^2 = Vi^2 + 2gx


    3. The attempt at a solution

    First I determined what the initial velocity was at the top of the window:

    X = Vi (time) + .5gt^2
    2.2 = Vi (.33) + .5 (-9.8) (.33)^2
    Vi = 8.28 m/s

    Then I used this velocity as my final velocity in this equation:


    Vf^2 = Vi^2 + 2gx
    8.28^2 = 0^2 + 2 (-9.8)x
    x = -3.5meters

    When I first inputted 3.5 meters into my online homework, it came out to be incorrect. I randomly decided to add -3.5 meters to 2.2 meters and the answer: 1.3 Meters, came out to be correct.

    I don't understand why I am adding -3.5 to 2.2.

    Thank you in advance.
     
  2. jcsd
  3. Sep 30, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Check your calculation of Vi; I'm seeing a lower value than 8.28 m/s. In particular, make sure the sign that you're ascribing to the acceleration constant matches your choice of coordinate system that the rest of the equation is assuming.
     
  4. Sep 30, 2014 #3
    I thought the acceleration due to gravity should always remain negative? With a positive acceleration, the Vi changes to 5.049.

    By using the positive acceleration, I got 1.3 Meters exactly!

    How exactly do you decide when to use a positive or negative acceleration due to gravity?
     
  5. Sep 30, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You chose the downward direction positive when you calculated with positive displacement ##\Delta x =2.2 m ##. If downward is positive, the downward gravitational acceleration is also positive.

    ehild
     
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