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Formula for free-falling object?

  1. Sep 13, 2011 #1
    Hi there, my question has to do with the velocity of a free falling object. I know that V=Vo + at, but if you dont know the time and the final velocity of the object, can you use another equation? I found this one"V^2=Vo^2 + 2(as)"-s being the distance travelled. Im trying to find the final velocity of an object that fell 134 meters from a bridge. Can somebody tell me please, if this is right, or point me in the right direction?

    Distance=134m
    Acceleration=9.8m.s^-2

    So I did:
    V^2=Vo^2 + 2(as)
    = V^2=0^2 + 2(9.8 x 134)
    =V^2=2(1313.2
    =V^2=2626.4
    v=51.25 m per second
    also i need to know the time, so i did:
    t=(v-vo)/a
    t=(51.25)/9.8
    t= 5.23 seconds
    im just not quite sure..also, is this equation for the average velocity or for the final velocity? thanks!!
     
  2. jcsd
  3. Sep 13, 2011 #2

    Hootenanny

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    Spot on. :approve:

    The final equation you used is formally the average velocity - but it is fine to use in problems where the acceleration is constant.
     
  4. Sep 13, 2011 #3
    oh cool! Thanks:) Also, the rest of the question was based on the kinetic energy and the momentum. I havent really done a lot of that so far, I have been trying to figure out velocity and acceleration equations! I think I have the right formulas, im just not sure how to convert my final answer.
    3. "Calculate the momentum of the person just before they hit the water" so if p=mv, then is the equation simply :
    p=mass(which is 70kg)x final velocity(51.25)
    p=70kg x 51.25m/s^-1
    and the to get the answer in kilograms per meter per second, i divided by 51.25 to get
    p=1.366kg.m.s^-1
    is this the right format to write an answer where there are two units of measurement?

    the last part of the question was:
    4. "Calculate the kinetic energy of the person, just before they hit the water"
    i found a formula which was KE=1/2(mv^2)
    so i did KE= 1/2(70kg x 51.25^2)
    KE=1/2(70kg x 2626.53m.s^-1)
    KE=35kg x 1313.265m.s^1
    and then i divided by 1313.265, so that i could get it in kilograms per meter per second, and my final answer was:
    KE= 0.02665kg.m.s^-1
    KE= 2.665 x 10^-2 kg.m.s^-1
    soo, not sure if that was all jibberish or not lol, please get back to meee:)
     
  5. Sep 13, 2011 #4

    Hootenanny

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    This is correct.
    This is also correct.

    There is no need to do additional conversions or calculations. You're quantities are already in appropriate (S.I.) units: kg.m/s for momentum and kg.m2/s2 = Joules for kinetic energy.
     
  6. Sep 13, 2011 #5
    Thanks:) :)
     
  7. Sep 13, 2011 #6

    Hootenanny

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    A pleasure :smile:
     
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