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Math Help - Quadric question

  1. #1
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    Quadric question

    a model rocket is launched from the roof of a bulinding. its flight path through air is modelled bu h=-5t^2+30t+10 where h is height of the rocket above ground in thr metres and t is the time after the launch in seconds.

    (a) when will teh rocket hit the ground?
    (b) when is the rocket 45m off the ground?
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  2. #2
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    Hello, usm_67!

    A model rocket is launched from the roof of a building.
    Its flight path through air is modelled by h = -5t^2+30t+10
    where h is height of the rocket above ground in metres and t is the time after the launch in seconds.

    (a) When will the rocket hit the ground?
    When the rocket hits the ground, h = 0.

    Solve: . -5t^2 + 30t + 10 \:=\:0

    Divide by -5: . t^2 - 6t - 2\:=\:0

    Quadratic Formula: . t\:=\:\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(-2)}}{2(1)}\;=\;\frac{6\pm\sqrt{44}}{2}

    Therefore: . t\;=\;3 + \sqrt{11}\;\approx\;6.3 seconds.


    (b) When is the rocket 45m off the ground?
    When is h = 45\;?

    Solve: . -5t^2 + 30t + 10\;=\;45

    We have: . -5t^2 + 30t - 35\;=\;0\quad\Rightarrow\quad t^2 - 6t + 7\;=\;0

    Quadratic Formula: . t\;=\;\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(7)}}{2(1)}\;=\;\frac{6 \pm\sqrt{8}}{2}

    Therefore: . t\;=\;3 + \sqrt{3}\;\approx\;4.4 seconds.
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  3. #3
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    do we have to always divide by common factor in this case dat is 5 but we r not using 5 anywhere and one more thing if we change this to in vertex form than how can we find the solution for a and b.
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  4. #4
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    Quote Originally Posted by usm_67
    do we have to always divide by common factor in this case dat is 5 but we r not using 5 anywhere and one more thing if we change this to in vertex form than how can we find the solution for a and b.
    Hello,

    I haven't got much time. So I will only show you step by step how you can get the vertex form of your parabol:

    h = -5t^2+30t+10

    h=-5(t^2-6t-2)=-5(t^2-6t+9-9-2) = -5((t-3)^2-11)

    h=-5(t-3)^2+55

    So the vertex is V(3,55). That means after 3 seconds the rocket had reached it's greatest height: 55 m.

    Greetings

    EB
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  5. #5
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    k thanks i got it
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  6. #6
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    Quote Originally Posted by Soroban
    Quadratic Formula: . t\;=\;\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(7)}}{2(1)}\;=\;\frac{6 \pm\sqrt{8}}{2}

    Therefore: . t\;=\;3 + \sqrt{3}\;\approx\;4.4 seconds.
    I didn't check through all of this (as my fancy calculator is still packed and I'm too lazy to unpack it! ), but I see two errors.

    The first is that \frac{6 \pm\sqrt{8}}{2} = \frac{6 \pm 2 \sqrt{2}}{2} = 3 \pm \sqrt 2.

    The second error is in using only the "+" sign. The initial height of the rocket is 10 m and the maximum height of the rocket is 55 m. Thus there will be TWO times that the rocket is at 45 m. The first is at t = 3 - \sqrt{2} s (on it's way up) and the second is at t = 3 + \sqrt{2} s (on it's way back down.)

    -Dan
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  7. #7
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    Hello, topsquark!

    . . . but I see two errors.
    Guilty as charged!

    The first is a simple typo . . . *blush*

    The second is a really stupid oversight . . . (*blush*)
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban
    Hello, topsquark!


    Guilty as charged!

    The first is a simple typo . . . *blush*

    The second is a really stupid oversight . . . (*blush*)
    Nothing stupid about it. I've been catching my students on that one for YEARS!

    -Dan
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