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Homework Help: An elevator moving up and a bolt falling down

  1. Aug 28, 2007 #1
    A bolt comes loose from the bottom of an elevator that is moving upward at a speed of 6.38 m/s. The bolt reaches the bottom of the elevator shaft in 3.85 s. How high up was the elevator when the bolt came loose?

    I integrated 6.38 to get the position equation and plugged in 3.85. However, I'm not really sue that's right at all. Could someone show me what type of equation I should be using?
  2. jcsd
  3. Aug 28, 2007 #2
    Should I get the height of the elevator at 3.85 seconds. Then use 9.8m/s and create a position equation for the falling bolt.
  4. Aug 28, 2007 #3
    firstly you have posted this in the wrong forum

    here are some hints which maybe able to help you

    [tex]s = ut + \frac{1}{2}at^2 [/tex]

    the velocity of the elevator is acting in the opposite direction to gravity, therefore it is a negative velocity in respect to gravity so gravity will be 9.8m/s/s and the elevator velocity will be -6.8m/s

    you have the value of t so use the equation above, where s is the displacement, i.e., how high up the elevator was in the shaft
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