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Math Help - Help with finding quadratic equation.

  1. #1
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    Help with finding quadratic equation.

    hello, totally new here
    and i'm having trouble deriving the equation out of this question...
    so here's the question:
    Alex was practising his 10m platform dive (i'm assuming that it means, he's above the water by 10m?). Because of gravity, the relation between his height, h, in metres, and the time, t, in seconds, after he dives is quadratic. If Alex reached a maximum height of 11.225m after 0.5 seconds, how long was he above the water after he dove?

    So.. i think i'm supposed to find the 2 roots to solve?

    thank you guys !
    Last edited by mr fantastic; December 11th 2010 at 07:50 PM. Reason: Title.
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  2. #2
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    The problem is says h is quadratic in t
    so h = At^2 + Bt + C

    First thing is to find A, B and C
    What do you know?
    In the beginning h = 10m and t = 0s
    when h = 11.225m, t = 0.5s

    Now you have 3 unknowns so there needs to be one more piece of information. I think that this requires knowing Earth's gravity is about 10 m/s^2 and the coefficient A is half of that so A = 5 m/s^2 (there should be an example in your text book).

    How much time was Alex above water?
    When h = 0 what is t?
    Remember t is always positive.
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  3. #3
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    Quote Originally Posted by Lyfee View Post
    hello, totally new here
    and i'm having trouble deriving the equation out of this question...
    so here's the question:
    Alex was practising his 10m platform dive (i'm assuming that it means, he's above the water by 10m?). Because of gravity, the relation between his height, h, in metres, and the time, t, in seconds, after he dives is quadratic. If Alex reached a maximum height of 11.225m after 0.5 seconds, how long was he above the water after he dove?

    So.. i think i'm supposed to find the 2 roots to solve?

    thank you guys !
    f(t)=at^2+bt+c

    t=0\Rightarrow\ f(0)=10\Rightarrow\ c=10

    The maximum occurs when the derivative is zero....

    2at+b=0

    2a(0.5)+b=0\Rightarrow\ a+b=0\Rightarrow\ b=-a

    f(t)=at^2-at+10

    \displaystyle\ f(0.5)=11.225\Rightarrow\frac{a}{4}-\frac{2a}{4}+10=11.225\Rightarrow\frac{a}{4}=-1.225

    a=-4.9

    f(t)=-4.9t^2+4.9t+10

    f(t)=0\Rightarrow\ 4.9t^2-4.9t-10=0

    This is zero after t=0 for

    t=\displaystyle\frac{4.9+\sqrt{4.9^2+40(4.9)}}{2(4  .9)}
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  4. #4
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    Quote Originally Posted by Lyfee View Post
    hello, totally new here
    and i'm having trouble deriving the equation out of this question...
    so here's the question:
    Alex was practising his 10m platform dive (i'm assuming that it means, he's above the water by 10m?). Because of gravity, the relation between his height, h, in metres, and the time, t, in seconds, after he dives is quadratic. If Alex reached a maximum height of 11.225m after 0.5 seconds, how long was he above the water after he dove?

    So.. i think i'm supposed to find the 2 roots to solve?

    thank you guys !
    This is easiest to solve using the Turning Point Form of the quadratic.

    You know that \displaystyle (x, y) = (0, 10) is the point where he begins, and \displaystyle (x, y) = (0.5, 11.225) is the maximum.


    So using the equation of the Turning Point Form of the quadratic:

    \displaystyle y = a(x - h)^2 + k, you can substitute your known values [ \displaystyle (h, k) is the turning point]

    \displaystyle 10 = a(0 - 0.5)^2 + 11.225

    \displaystyle 10 = a(-0.5)^2 + 11.225

    \displaystyle 10 = 0.25a + 11.225

    \displaystyle -1.1225 = 0.25a

    \displaystyle -4.49 = a.


    So the equation of your quadratic is

    \displaystyle y = -4.49(x - 0.5)^2 + 11.225.


    Now you are wanting to find out how long he is above the water - in other words, where he comes back to the ground, the \displaystyle x intercept.

    So make \displaystyle y = 0...

    \displaystyle 0 = -4.49(x - 0.5)^2 + 11.225

    \displaystyle -11.225 = -4.49(x - 0.5)^2

    \displaystyle 2.5 = (x - 0.5)^2

    \displaystyle \frac{5}{2} = \left(x - \frac{1}{2}\right)^2

    \displaystyle \frac{10}{4} = \left(x - \frac{1}{2}\right)^2

    \displaystyle \pm \frac{\sqrt{10}}{2} = x - \frac{1}{2}

    \displaystyle \frac{1 \pm \sqrt{10}}{2}= x.


    Obviously we can only accept the positive value, so he was in the air for \displaystyle \frac{1 + \sqrt{10}}{2} \approx 2.08114 seconds.
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  5. #5
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    Quote Originally Posted by snowtea View Post
    The problem is says h is quadratic in t
    so h = At^2 + Bt + C

    First thing is to find A, B and C
    What do you know?
    In the beginning h = 10m and t = 0s
    when h = 11.225m, t = 0.5s

    Now you have 3 unknowns so there needs to be one more piece of information. I think that this requires knowing Earth's gravity is about 10 m/s^2 and the coefficient A is half of that so A = 5 m/s^2 (there should be an example in your text book).

    How much time was Alex above water?
    When h = 0 what is t?
    Remember t is always positive.
    As another alternative,
    if you are to find a solution not involving calculus,
    you could also utilise the fact that the graph is symmetrical about the maximum.

    Hence....

    f(0)=f(1)

    as the maximum occurs at f(0.5).

    With c=10,

    you then have a pair of equations to solve for the unknowns "a" and "b"....

    f(0.5)=11.225\Rightarrow\ a(0.5)^2+b(0.5)+10=11.225

    f(1)=10\Rightarrow\ a+b+10=10

    Your choice of solution depends on the context of the question.

    Do you need to use the constant of gravity?

    It can be answered simply by solving a system of simultaneous equations.
    3 unknowns require 3 equations.

    f(t)=at^2+bt+c

    f(0)=c=10

    f(0.5)=\frac{1}{4}a+\frac{1}{2}b+10=11.225

    f(1)=f(0)\Rightarrow\ a+b+10=10

    Then find for t>0, f(t)=0
    Last edited by Archie Meade; December 12th 2010 at 11:01 AM.
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  6. #6
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    Whenever you know the Turning Point, the Turning Point Form of the graph is the easiest to work with...
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