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Math Help - Dropping object from height

  1. #1
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    Dropping object from height

    Hello,
    An object is released from a height of 25M. How far has the object travelled during the 3rd second of fall?

    I don't really know how to do this - what formula/'s should I use?

    thanks!
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  2. #2
    Senior Member vincisonfire's Avatar
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    Formula is
     h = v_0 \cdot t + a\frac{t^2}{2}
     0 = v_0
     a = g
     h =  g\frac{t^2}{2}
     -25 =  g\frac{t^2}{2}
    You can find that object touches the ground at 2.26 seconds
    Find how far it went at 2 seconds and do 25m minus this answer
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  3. #3
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    The formula is

    h = ut +\frac{1}{2}gt^2

    where h = height

    u = initial velocity (which is 0 here)

    g = accleration due to gravity = 9.81 \;m/s^2

    t = time.
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  4. #4
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    Got it - thanks a lot! Oddly worded question, that was the tough part.
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  5. #5
    Senior Member vincisonfire's Avatar
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    Don't you think the object should travel further during the 3rd second.
    It starts from heigth 25 m. And goes down 19 m in two seconds as you said.
    It is going to make the rest (6m) in the third second.
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  6. #6
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    Quote Originally Posted by vincisonfire View Post
    Don't you think the object should travel further during the 3rd second.
    It starts from heigth 25 m. And goes down 19 m in two seconds as you said.
    It is going to make the rest (6m) in the third second.
    Height fallen by object after 2 seconds is given by

    h = 0 + \frac{1}{2}(9.81)(2)^2

    h=19.62\;m

    And, Height fallen by object after 3 seconds is given by

    h = 0 + \frac{1}{2}(9.81)(3)^2

    h=44.15\;m>25\;m

    so, object touches ground before 3 seconds

    Height fallen by object in 3rd second = 25 - 19.62 = 5.38 m
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  7. #7
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    it only takes a bit over 2.2 seconds to hit the ground from a height of 25 m, so asking how far it travels in the third second is poorly posed.

    if it had enough height to fall the "third" second, then from t = 2 to t = 3 seconds ...

    \Delta y = v(2) \cdot \Delta t - \frac{1}{2}g(\Delta t)^2 = -19.6(1) - 4.9(1) = -24.5 \, \, m

    if the intent was to determine the distance it fell from t = 2 seconds until it hits the ground, ...

    \Delta y = -\frac{1}{2}g(2)^2 = -19.6 \, \, m

    leaving 5.4 m to fall from t = 2 to impact.
    Last edited by skeeter; October 23rd 2008 at 01:59 PM. Reason: fix bonehead mistake with a negative sign
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