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Free fall question

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    a ball is thrown straight down from the top of a 220 feet building with an initital velocity of -22ft/s. What is the velocity of the ball after 3 seconds? What is the velocity of the ball after falling 108 feet?


    2. Relevant equations
    s(t) = -16t^2 + Vi(t) + si vi = initial velocity, si = initial position


    3. The attempt at a solution
    s(t) = -16t^2 +Vi(t) +si
    = -16t^2 - 22ft/s + 200ft

    V(t) = -32(t) - 22ft/s
    V(3) = -96 ft/s - 22ft/s = -118 ft/s

    I can seem to figure out how to determine the velocity of the ball after it has dropped 108ft.

    I should be able to do this, but i'm missing something, somewhere on what should be pretty basic.

    Thanks for any guidance!
     
  2. jcsd
  3. Oct 25, 2008 #2

    nrqed

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    There are two ways to do it.

    The quickest way is to use [tex] V_f^2 = V_i^2 - 2 g \Delta y [/tex].

    The other ways is to use the equation [tex] y= V_i t -1/2 g t^2 [/tex] to first find the time to drop 108 feet and then to plug this into [tex] V_f = V_i - g t [/tex]
     
  4. Oct 25, 2008 #3
    Thanks for the reply. I have a solution that makes sense now. :)

    I appreciate your help!
     
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