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Free fall: total distance expressed in terms of distance fallen in nth second

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A body falls vertically from rest. During the nth second it falls a distance d. Prove that by the end of the nth second it has fallen a total distance (D) of (2d+g)^2/8g


    2. Relevant equations
    x = x0 + V0 + 1/2at^2
    where
    x0 = initial position
    v0 = initial velocity
    a = acceleration
    t = time


    3. The attempt at a solution
    The total distance is D = 1/2gn^2 because the object has fallen for n seconds. At the end of the first second d=D and t^2 = 2d/g. I'm unsure about how to proceed from here. Any help gratefully received...
     
  2. jcsd
  3. Dec 28, 2008 #2

    PhanthomJay

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    jemerlia, welcome to PF!

    Hint: What is the distance fallen after (n-1) seconds?
     
  4. Dec 28, 2008 #3
    Thanks for the hint - clearly the distance d is given by
    d = 1/2gn^2 - 1/2g(n-1)^2
    which (according to my rusty maths) simplifies
    = 1/2g(2n-1)

    It appears that t^2 = 2d/g is useful here but substitution appears to give nonsense. I 've obviously missed or misunderstood something.
     
  5. Dec 28, 2008 #4

    rl.bhat

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    {d = 1/2gn^2 - 1/2g(n-1)^2
    which (according to my rusty maths) simplifies
    = 1/2*g*(2n-1)}
    This is right. Now find n in terms of d and g.
    Now the total distance fallen in n seconds is D = 1/2*g*n^2. Substitute the value of n. You will get the required answer.
     
  6. Dec 29, 2008 #5
    Many thanks for the help during the "holidays"...

    Because d = 1/2 * g * (2n-1) then

    n = (2*d/g + 1) / 2

    = d/g + 1/2

    However, when n is substituted into

    D = 1/2*g*n^2

    as

    D = 1/2*g*(d/g + 1/2)^2

    it does not produce the expected result. I guess I have a problem with the arithmetic somewhere... advice gratefully received...
     
  7. Dec 29, 2008 #6

    rl.bhat

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    D = 1/2*g*(d/g + 1/2)^2

    D = 1/2*g*(2d + g)^2*1/4g^2
    = (2d + g )^2/8g
     
  8. Dec 29, 2008 #7
    Thank you to everyone who gave their time to help me with this problem. It has served to identify the areas I must work on.
     
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