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Amusment park yesterday for physics class

  1. May 12, 2004 #1
    When to an amusment park yesterday for physics class. Went on a ride called the "hellevator" where you sit down and it shoots up a certain distance then lets you free fall. My question is if the acceleration upwards occurs in the first 5 meters how do you figure out that acceleration?
  2. jcsd
  3. May 13, 2004 #2


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    You are going to need more information to find the acceleration.

    Given an acceleration of a m/s2, the speed after t seconds is v= at and the height is h= (1/2)at2.

    If you know the time, T, taken to get to that first 5 meters, you could solve
    5= (1/2)a(T2): a= 10/T2.

    After that first time, T, you continue upwards under the influence of grativity:
    you speed at time t will be v= -g(t-T)+ aT and height will be h= 5+aTt- (1/2)g(t-T)2.

    At the maximum height, v= -g(t-T)+ aT= 0 so you can solve for t (in terms of T). If you know that maximum height (usually easier to measure than time), you could solve the second equation for a.
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