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March 26, 2017

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Carol drops a stone into a mine shaft 12.6 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft?

  • Physics - ,

    distance = 1/2 g*t^2

  • Physics - ,

    The mine shaft is 122.6 m deep.

    How do I find time?

  • Physics - ,

    The mine shaft is 122.6 m deep.

    How do I find time?

  • Physics - ,

    The mine shaft is 122.6 m deep.

    How do I find time?


    you have the distance so..

    12.6m or 122.6m (you changed it or you had a typo) but the set up is the same.

    the equation Dr.Bob gave was correct but to elaborate on it:

    Y(final height)= Y(initial height) + v(initial)t + 1/2 a(acceleration) t^2

    is where he got the
    distance= 1/2g*t^2

    so since acceleration in y direction (or toward earth) is 9.81m/s^2 you use that as g or gravity in Dr.Bob's given equation or as "a" in my equation.

    so now plugging in...(you only have to rearrange the equation to solve for t.)

    sqrt[2(distance)/g] = t

  • Physics - ,

    Y(final height)= Y(initial height) + v(velocity initial)t + 1/2 a(acceleration) t^2

    is where he got the
    distance= 1/2g*t^2

    I forgot to mention that the reason tha the 1/2g*t^2 does NOT include the initial velocity in my equation is because you assume that the object that is being dropped is at rest before it is dropped and thus the initial velocity is 0 and thus the 2nd part of the equation is canceled out because it is 0

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