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A falling object's speed

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A 69 kilogram person falls from a 35 meter tower head first to the ground

    2. Relevant equations

    What is the maximum velicoity he reaches before impact?

    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2009 #2

    DaveC426913

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    See that #3? That's where you make an attempt at a solution.

    We don't spoon-feed homework here.
     
  4. Nov 3, 2009 #3
    Hint: Use your kinematic equations for uniform acceleration.
     
  5. Nov 3, 2009 #4
    Sorry if I am a dumbum. I'm 57 and have forgotten all my equations, not that I was very good at physics. I am doing research for a stuntman who has survived this fall and want to know what speed he reached.
     
    Last edited by a moderator: Nov 4, 2009
  6. Nov 3, 2009 #5

    DaveC426913

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    The https://www.physicsforums.com/showthread.php?t=5374" that you agreed to when you signed up require you to attempt a solution:

    "NOTE: You MUST show that you have attempted to answer your question in order to receive help."


    Furthermore, we are expressly requested not to do work for people here; this is afterall, and homework help forum.
     
    Last edited by a moderator: Apr 24, 2017
  7. Nov 3, 2009 #6

    DaveC426913

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    V = Vo + at
    X - Xo = Vot + .5at2
    v2 = vo2 + 2a(X - Xo)
    X - Xo = .5(Vo + V)t

    V is final velocity in units of meters per seconds (m/s)
    Vo is initial velocity in units of meters per seconds (m/s)
    a is acceleration in units of meters per squared seconds (m/s2)
    t is time in seconds (s)
    X is final displacement in units of meters (m)
    Xo is initial displacement in units of meters (m)
     
  8. Nov 3, 2009 #7

    Integral

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    The mass does not matter, any falling object will have the same speed after falling 35m.

    First you need the time of the fall.

    x = fall distance = 35m
    a = acceleration due to gravity = 9.8 m/s
    t = time of fall.

    x = .5 a t^2
    [tex] t = \sqrt { \frac {2x} a } [/tex]

    Given the time of the fall the speed is

    v = at

    so [tex] v = \sqrt {2 a x} [/tex]

    That should give you all the equations you need. can you take it from here?
     
  9. Nov 3, 2009 #8
    thank you DaveC426913 and Integral. I am new to this board and appreciate your help.
    I will not bother you again.
     
  10. Nov 3, 2009 #9

    DaveC426913

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    Feel free to bother. That's what we're here for. :wink: But there's rules is all.
     
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