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Math Help - Distance and Velocity

  1. #1
    Member Veronica1999's Avatar
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    Distance and Velocity

    A stone is dropped into a well and the report of the stone striking the bottom is heard 7.7 seconds after it is dropped. Assume that the stone falls 16t squared feet in t seconds and that the velocity of sound is 1,120 feet per second. The depth of the well is:

    Help pls.
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  2. #2
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    How can something fall an amount of squared feet per unit time? Falling = height = 1 dimension.
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  3. #3
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    Quote Originally Posted by Veronica1999 View Post
    A stone is dropped into a well and the report of the stone striking the bottom is heard 7.7 seconds after it is dropped. Assume that the stone falls 16 (t squared) feet in t seconds and that the velocity of sound is 1,120 feet per second. The depth of the well is:

    Help pls.
    1. Per definition distance = speed \cdot time.

    2. The total time is split into two parts: The time t which is needed by the falling stone and the remaining time (7.7 - t) which is needed by the sound. The falling stone and the sound have to pass the same distance d:

    d_{stone} = 16 \cdot t^2

    d_{sound} = 1120 \cdot (7.7-t)

    3. Solve for t:

    16 \cdot t^2 = 1120 \cdot (7.7-t)

    4. Plug in the value of t into one of the equations determining d to get the depth of the well.
    Spoiler:
    I've got 784 '
    Last edited by earboth; July 11th 2010 at 10:51 PM.
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  4. #4
    Member Veronica1999's Avatar
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    Thank you. Thank you. Thank you.
    Everything is so clear now.
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