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Math Help - Using the quadratic formula

  1. #1
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    Using the quadratic formula

    I have been asked to calculate the maximum height a ball can achieve after being lauched, using;

    h = - 4.9t^2 + 19.6t - 12.6

    I knew straight away I required the time, so I did the following;

    t = -b + or - square root b^2 - 4ac / 2a

    t = - 19.6 + or - square root 19.6^2 - 4 (- 4.9)(- 12.6)
    2 x ( - 4.9)

    t = - 19.6 + or - square root 384.16 - 246.96
    2 x ( - 4.9)

    t = - 19.6 + or - square root 137.20
    2 x( - 9.80)

    t = - 19.6 + or - 11.71
    - 9.80

    t = - 7.89
    - 9.80

    t1 = 0.81

    t2 = - 19.6 + or - 11.71
    - 9.80

    t2 = - 31.31
    - 9.80

    t2 = 3.20

    I then used both routes to conclude the height in metres to determine if there was any significant difference, there wasn't so I will use t2;

    h = - 4.9 x 3.20^2 + 19.6 x 3.20 - 12.6

    h = - 50.18 + 62.72 - 12.6

    h = 0

    This is the third answer I have had using the above formula?

    I have had first 65m, then 34m and now 0 metres?

    I can see I must have tapped a interger in wrong in the calculator or not used it correctly, but can I ask, does the above now look like I have got it right?

    Thanks

    David
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Using the quadratic formula

    There's been asked the maximum height the ball can achieve, you're calculating the time wherefore the h=0 and so when the ball reach the ground.
    Use:
    t=\frac{-b}{2a}

    Now you're calculating after how much time the ball will reach his highest point and then you've just to enter this time t in the equation.
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  3. #3
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    Re: Using the quadratic formula

    Quote Originally Posted by Siron View Post
    There's been asked the maximum height the ball can achieve, you're calculating the time wherefore the h=0 and so when the ball reach the ground.
    Use:
    t=\frac{-b}{2a}

    Now you're calculating after how much time the ball will reach his highest point and then you've just to enter this time t in the equation.
    Hi Siron, so are you saying that the height above the cliff is 7m using the equation you quote?

    I see what you mean now, the equation you use is taking the average of the two routes of the t1 and t2 I found, I never thought about averaging the routes?

    Thanks

    David
    Last edited by David Green; July 25th 2011 at 12:36 PM. Reason: additional information
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Using the quadratic formula

    Yes, the maximum height a ball can achieve is indeed 7m en that's you wanted to calculate. If you want to calculate after how much time the ball will reach the ground and so h=0, then you've indeed to use the quadratic formula.
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  5. #5
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    Re: Using the quadratic formula

    Quote Originally Posted by Siron View Post
    Yes, the maximum height a ball can achieve is indeed 7m en that's you wanted to calculate. If you want to calculate after how much time the ball will reach the ground and so h=0, then you've indeed to use the quadratic formula.
    Thanks for your help, much appreciated.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Using the quadratic formula

    You're welcome!
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  7. #7
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    Re: Using the quadratic formula

    Or you can use the completing the square method to find the maximum height and time taken to reach maximum height

    h = - 4.9t^2 + 19.6t - 12.6

    = -4.9 (t^2 - 4t) - 12.6

    = -4.9 (t^2 - 4t + 4 - 4 ) - 12.6 Add and subtract square of half the coefficient of linear term to complete the square.

    = -4.9(t - 2)^2 + 19.6 - 12.6

    h = -4.9(t - 2)^2 + 7

    Hence, the maximum height is 7 and after 2 second, which can be found from the vertex (2, 7) of the parabola.
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Using the quadratic formula

    True, there are a lot of methods to calculate the maximum/minimum of a parabola. Calculating the first derivative and then f'(x)=0 is also a possibilty (but David hasn't learned derivatives yet).
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