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Math Help - Question about Quadratic equation/parabola/factoring stuff

  1. #1
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    Question about Quadratic equation/parabola/factoring stuff

    The Question says:
    A model rocket is shot into the air and its ath is approximated by the relation h= -5t + 5t + 30 where h is the height above the ground in metres and t is the time elapsed in seconds.
    a) How tall is the building?
    This is what I did:
    h=-5t+5t+30
    h=-5(t^2-t-6)
    h=-5(t-2)(t+3)

    sub t=0

    h= -5(t-2)(t+3)
    h= -5(0-2)(0+3)
    h= -5(-2)(3)
    h= -5(-6)
    h=30

    B) WHEN WILL THE BALL HIT THE GROUND?

    I thought I could sub h=0 into the equation for this so I did it as follows:

    h= -5t(t-2)(t+3)
    0= -5t(t-2)(t+3)
    0= -5(t^2+t-6)
    0= -5t^2 -5t + 30

    I don't know where to go from here or how to finish the problem or if I did anything wrong. Please help.
    Last edited by MathGuru; November 30th 2005 at 04:39 PM.
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  2. #2
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    You have the relation for the height, so set the height equal to zero and solve for the time it will take to get there. There will be two zeroes of course, but I think you can figure out which one is relevant.
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  3. #3
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    For the height of the building it was not necessary to factorize. Just plug t=0
    h=-5t^2+5t+30 -> h(0)=-5*0^2+5*0+30=30.

    Now, it was not absolutely unecessary since you showed that you know how to factorize. There is only a little sign mistake i.e. when you look for your two numbers a*b=-6 and a+b=-1 it gives you -3 and 2 (not 3 and -2). It is a common mistake so you should check always your answer.

    h=-5t^2+5t+30=-5(t^2-t-6)=-5(t-3)(t+2).

    So now we look for t when h=0.
    We don't use our equation in the form : 0=-5t^2+5t+30 since we can not isolate t so what we do is the factorization you did : 0=-5(t-3)(t+2).
    Why do we factorize? Because since you have "something1"*"something2"=0 so either (or both) these "something"s is =0. Logic ! So you factorize when one of the two sides =0.

    Ok so either -5=0 (does not make sense) or (t-3)=0 or (t+2)=0.

    Now, when does t-3=0 ? Well it is when t=3s (when you isolate t).
    And t+2=0 when t=-2s. Of course -2 does not make sense in our problem so you take 3s.

    Ok? Now, if you want to understant why are there two answers when only one is right. It is because the equation h=-5t^2+5t+30 is in the math world (so it just a parabola which has normally 2 zeros - see figure) since the x axis has a negative part. Now putting x as time and y as height makes x >=0 since there is no t<0. That is why we just consider the t=3s as a solution.
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    Last edited by hemza; December 1st 2005 at 05:57 AM.
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  4. #4
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    right, that makes a lot of sense, hemza, Thank-you. My teacher also said that that was right.
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