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Math Help - Equation of motion help please.

  1. #1
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    Equation of motion help please.

    A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls half the height of the cliff in the last three seconds of its fall. What is the total fall time? (use g = 9.81 m/s^2)

    This is my attempt, since the rock is thrown horizontally it makes no difference to the fall time.

    so:

    \frac{3}{\left(1-\sqrt1-0.5\right)}=10.24264\:secs

    Can you please confirm?
    Last edited by 1LISTEN; December 28th 2009 at 03:59 PM.
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  2. #2
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    Can any one help please? i'm convinced my time is correct.

    \frac{1}{2}at^2 = 4.905\times7.24264^2 = 257.295\:m

    \frac{1}{2}at^2 = 4.905\times10.24264^2 = 514.59\:m

    \frac{257.295}{514.95}= 0.5
    Last edited by 1LISTEN; December 28th 2009 at 03:29 PM.
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  3. #3
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    Hello, 1LISTEN!

    A rock is thrown horizontally from a cliff with a speed of 15 m/s.
    It falls half the height of the cliff in the last three seconds of its fall.
    What is the total fall time? .(Use g = 9.81 m/sē) .
    Also not needed

    Since the rock is thrown horizontally its initial speed
    makes no difference to the fall time. . . . . . Right!

    The height of an dropped object is: . y(t) \:=\:h_o - \frac{1}{2}gt^2
    . . where h_o = initial height.


    In the first T seconds, the rock falls half the height: . y(T) \,=\,\frac{1}{2}h_o
    . . h_o - \frac{1}{2}gT^2 \:=\:\frac{1}{2}h_o \quad\Rightarrow\quad h_o \:=\:gT^2 .[1]

    In the next 3 seconds, the rock strikes the ground: . y(T+3) \,=\,0
    . . h_o - \frac{1}{2}g(T+3)^2 \:=\:0 \quad\Rightarrow\quad h_o \:=\:\frac{1}{2}g(T+3)^2 .[2]


    Equate [1] and [2]: . gT^2 \:=\:\frac{1}{2}g(T+3)^2 \quad\Rightarrow\quad T^2 - 6T - 9 \:=\:0

    Quadratic Formula: . T \;=\;\frac{6\pm\sqrt{72}}{2} \;=\;3 \pm3\sqrt{2}


    Therefore, the total fall time is: . T + 3 \;=\;6 + 3\sqrt{2} seconds.

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