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Math Help - Quadratic Forumla: Dropping an object

  1. #1
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    Quadratic Forumla: Dropping an object

    A bit of a repost, but I have a different question. This is the question given:

    If an object is thrown downward at 26 m/s from the top of a building, 102m high, the equation relating to displacement (meters) and time (seconds) is:

    ]s(t) = 1/2 gt^2 + 26t + 102

    Find the time for an object to reach the ground from the top. Use g=-9.81

    So I have come up with

    (-26) +/- \sqrt (26)^2 - 4(-4.905)(102) / 2(-4.905)

    Unfortunately I keep getting 2 negative solutions.

    I was given this explaination:


    The initial velocity is (you are working with s positive upwards) and , so your equation for displacement is:



    which I can tell you has exactly one positive root (and in consequence one negative root, but that is not the one you want).

    --------------------

    My question is, how do I figure out I need to use -26t? Am I supposed to look at the given formula and change 26t to a negative number because I am dropping an object?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Quadratic Forumla: Dropping an object

    Quote Originally Posted by polskon View Post

    My question is, how do I figure out I need to use -26t? Am I supposed to look at the given formula and change 26t to a negative number because I am dropping an object?
    It depends on how you choose your frame of reference ...
    In this case they have chosen the 'y-axis' upwards and because your acceleration and velocity are orientated downwards (in this case) they get a negative sign.
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  3. #3
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    Re: Quadratic Forumla: Dropping an object

    Id just like to update: My problem with the formula was that I kept getting 2 negative solutions. To my surprise, it was my fault. I was continuously inputting the answers wrong on my calculator. I guess I didn't understand the formula 100%. Thanks!
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