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Math Help - A ball is thrown up at the edge of a cliff....

  1. #1
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    A ball is thrown up at the edge of a cliff....

    A ball is thrown up at the edge of a 298 foot cliff. The ball is thrown up with an initial velocity of 72 feet per second.
    Its height measured in feet is given in terms of time t, measured in seconds by the equation h=-16t^2+72t+298.
    a. How high will the ball go and how long does it takes to reach that height? __________ feet, ___________ seconds
    b. How long does it take the ball to come back to the ground ? ___________ seconds

    I have part (a) but I'm having trouble answering part (b)
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: A ball is thrown up at the edge of a cliff....

    Set h = 0 and solve for t, discarding the negative root.
    Thanks from spyder12
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    Re: A ball is thrown up at the edge of a cliff....

    I have a very similar problem and I'm stuck on finding the final velocity when the ball hits the ground, and I was wondering if I could tack it onto this thread. s(t) = 48 + 48 t - 16 t^2. If I set s(t) to 0 and use the quadratic formula I get a \frac{3+ or-sqrt(-3)}{2}, how do I disregard the complex number, or did I do something wrong?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: A ball is thrown up at the edge of a cliff....

    You have incorrectly applied the quadratic formula:

    First, put the quadratic into standard form:

    16t^2-48t-48=0

    Divide through by 16:

    t^2-3t-3=0

    Hence:

    t=\frac{3\pm\sqrt{(-3)^2-4(1)(-3)}}{2(1)}=\frac{3\pm\sqrt{21}}{2}

    Then discarding the negative root, we find:

    t=\frac{3+\sqrt{21}}{2}
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    Re: A ball is thrown up at the edge of a cliff....

    Thanks, I didn't even realize I had evaluate it wrong.
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