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Falling object - relating distance interval to time interval

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data

    An object falls from the top of a 103.6m high building. The vertical
    distance between two consecutive windows A and B is 4.0m. The first window/storey is
    also 4m from the top of the building. The object reaches window B 0.1s after it has reached window A.

    Which storey is A ?

    - Top of building, 103.6m above ground
    |
    | 4m
    |
    = Window
    |
    | 4m
    |
    = Window
    |
    |4m
    |
    = Window
    .
    .
    = Window A
    |
    |4m,time to fall from A to B is 0.1s
    |
    = Window B
    |
    |
    |
    .
    .
    _ Ground


    2. Relevant equations
    d= 1/2gt2


    3. The attempt at a solution
    The height of the building seems irrelevant (it is not a multiple of 4) except to indicate the object will be falling quickly enough to satisfy the conditions of the problem. What appears to be required is to relate the time interval (0.1s) to the distance interval 4n and 4n+1

    The time for an object to fall a given distance from a start velocity of 0m.s will be
    SQRT(2D/g)

    The most promising approach appeared to be to relate the distance between the windows A and B expressed as 4n metres to the time interval and solve for n (the number of storeys from the top) so that:
    time interval = time to reach A - time to reach B

    Letting DB = distance from top to Window B, DA=distance from top to Window A

    0.1 = SQRT(2DB/g) - SQRT(2DA/g)

    Substituting DB = 4(n+1) metres, DA = 4n metres

    0.1 = SQRT(2(4(n+1))/g) - SQRT(2(4n)/g)

    Sadly I'm in error because the n's cancel!!!

    What have I done wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    0.1 = SQRT(2(4(n+1))/g) - SQRT(2(4n)/g)
    Wright this equation as
    0.1 + SQRT(2(4n)/g) = SQRT(2(4(n+1))/g).
    Square both sides and solve for n.
    From the height of building it is clear the last window from the top must be 25th.
     
  4. Jan 30, 2009 #3
    Still puzzled...
    n should give the number of storeys A is from the top...

    Following your suggestion, I have:
    0.12+ (8n)/g = (8n+8)/g

    then
    0.01g + 8n = 8n +8

    The n's still cancel - I've obviously done something wrong

    The expected answer is 6
     
  5. Jan 30, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    0.1^2+ (8n)/g = (8n+8)/g
    This step is wrong.
    It should be
    0.1^2+ (8n)/g + 2*0.1*(8n/g)^1/2= (8n+8)/g
     
  6. Jan 30, 2009 #5
    Please check my working - I don't get the specified answer - but it's possible that answer counted the storeys from the ground up...

    I now see my mistake in squaring one side of the expression... starting from there:

    0.1^2+8n/g+(0.2 x SQRT(8n/g)) = (8n+8)/g

    Multiply both sides by g

    0.01g +8n +(0.2g x SQRT(8n/g)) = 8n + 8

    Cancel the 8n terms

    0.01g + (0.2g x SQRT(8n/g)) = 8

    Re-arrange:

    0.2g x SQRT(8n/g) = 8-0.01g

    SQRT(8n/g) = (8-0.01g)/0.2g

    8n/g = ((8-0.01g)/0.2g)^2

    n = (g x ((8-0.01g)/0.2g)^2)/8

    Taking g to be 9.81ms^-2

    Gives n as 19.91 (4sfs) so I guess n = 20
     
  7. Jan 30, 2009 #6

    rl.bhat

    User Avatar
    Homework Helper

    Yes. n is correct.
    As I told you before, from the height of building (103.6 m) it is clear the last window from the top must be 25th. Upto A number of windows 20. So below A there are 5 windows. Hence A must be 6th storey.
     
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