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Math Help - Word problem

  1. #1
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    Word problem

    I don't even know how to begin this.

    You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 7 seconds. The depth of the well is _____ feet. Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.
    Note: After seconds the rock has reached a depth of feet where d=16t^2
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  2. #2
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    Quote Originally Posted by md56 View Post
    I don't even know how to begin this.

    You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 7 seconds. The depth of the well is _____ feet. Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.
    Note: After seconds the rock has reached a depth of feet where d=16t^2
    t_1 = time down

    t_2 = time for the splash sound to return

    * t_1 + t_2 = 7



    d = 16t_1^2

    d = 1100t_2

    * 16t_1^2 = 1100t_2

    solve the system (*) for t_1 or t_2 ... then find d
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  3. #3
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    2/1 following 16t an exponent?
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  4. #4
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    Quote Originally Posted by md56 View Post
    2/1 following 16t an exponent?
    16(t_1)^2 ... the 1 is a subscipt.
    Last edited by skeeter; October 1st 2009 at 05:11 PM. Reason: fix typo
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  5. #5
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    now I am lost.
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  6. #6
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    Quote Originally Posted by md56 View Post
    now I am lost.
    my typo ... should be 16(t_1)^2 ... put the opening parenthesis too soon.
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  7. #7
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    I don't know what is required . . . let's find SOME of the unknown.

    I will use the data above,

    T + t = 7 - - - - (1)

    Let T = time going down, in seconds

    t = time going up, in seconds

    d = depth of the well, feet

    d = (acceleration of the rock)(Time going down)^2 = 16T^2 - - - - (2)

    also, d = (speed of sound)(time going up) = 1100t - - - - (3)

    equate (2) and (3)

    16T^2 = 1100t

    t = 16T^2/1100 - - - - (4)

    substitute (4) in (1),

    T + t = 7, that is t = 16T^2/1100

    T + 16T^2/1100 = 7, cross-multiply

    1100T + 16T^2 = 7700,rearrange them

    16T^2 + 1100T - 7700 = 0,

    it is now quadratic in T, can you solve it now?





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  8. #8
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    I am getting 6.403, and it is incorrect
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  9. #9
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    Quote Originally Posted by md56 View Post
    I am getting 6.403, and it is incorrect
    6.403 is the time it takes for the rock to get to the bottom after being dropped.

    re-read the question ... what is it asking for?
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  10. #10
    Senior Member pacman's Avatar
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    T = (5/8)(3(sqrt 473) -55) = 6.4 sec,

    T + t = 7,

    t = 7 - 6.4 = 0.6 sec.

    depth of the well = d = 1100t = 1100(0.6) = 660 feet.

    Wow! Is not that too deep for a well?
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