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The instantaneous velocity of a freely falling object

  1. Nov 13, 2008 #1
    What is the instantaneous velocity of a freely falling object 10 s after it is released from a position of rest? What is its average velocity during this 10 s interval? How far will it fall during this time?

    I had asked a few friends for help and I keep getting mixed answers.

    v = gt
    98m/s = 9.8m/s * 10
    Average V = initial v + final v
    2
    Average V = 0m/s + 98m/s
    2

    Average V=49m/s
    Distance Traveled d = ½ gt squared
    ½ * 9.8m/s * 10 squared = 490m

    or

    Vf=Vo+at = 0msec+10msec2(10sec)= 100msec

    AverageVelocity=Vf+Vo2=100ms+0ms2=50msec

    X=Vot +12at2=0ms(10sec)+12(10msec2)(10sec)2=500 meters

    Who's on the right path?
     
  2. jcsd
  3. Nov 13, 2008 #2

    LowlyPion

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    Welcome to PF.

    You can approach average velocity either way.

    V = a*t => Vavg = 1/2*a*t

    or X = 1/2*a*t2

    Vavg = Xtot/Ttot = 1/2*a*t2/t = 1/2*a*t

    V2 = 2*a*x = 2*a*(1/2*a*t2) = a2t2

    V = a*t

    They are all interrelated.
     
  4. Nov 13, 2008 #3
    The answers came up slightly different would it matter?
     
  5. Nov 13, 2008 #4

    LowlyPion

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    There is no difference that I see. Acceleration is due to gravity in both cases.

    The bottom equation uses g = 10 instead of 9.8.
     
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