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    talar's Avatar
    talar Posts: 1, Reputation: 1
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    #1

    Aug 8, 2008, 09:55 AM
    Quadratic Function Solving
    The height of a flare, h metres, above the release position, after t seconds, is h= -5t sqaured + 100t

    A) What is the height of the flare after 3s
    B) What is the maximum height reached by the flare?
    C) What is the height of the flare after 25s
    D) Does your answer in part c make sense. Explain
    E) Determine algebraically the time for which the flare is higher than 80m

    PLEAASE HELP ME NOWWW
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Aug 8, 2008, 03:33 PM
    This is very straighforward.



    part a: What is the height after 3 seconds? Think a little about it. If t is time and y is the height, wouldn't you just plug in t=3?

    Did you graph this? The max height is the vertex of the concave down parabola.

    You can find the vertex by using , then plug that value into your given quadratic to find the y-value which is the max height.

    For part c, what do you get when you enter t=25 into your quadratic? Does it make sense?

    If the ground is at y=0, then does a negative value make any sense? That would be below the ground wouldn't it?

    To find when it is higher than 80 m, set y=80 and solve for t. You should get two values. One for on the way up and one for on its way down.
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