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Math Help - Quadratic Model

  1. #1
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    Quadratic Model

    For this function h(t)=10-4t^2 how do you determine what time interval the ball will be more than 40 feet in the air?
    Last edited by rebak; February 16th 2009 at 10:17 PM.
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  2. #2
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    Quote Originally Posted by rebak View Post
    For this function h(t)=10-4t^2 how do you determine what time interval the ball will be more than 40 feet in the air?
    There must be a typo!

    The graph of the given function is a parabola opening down. The vertex of the parabola (that's the highest point the ball can reach!) is at V(0, 10).

    Thus: The ball never reaches a height of 40' .
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  3. #3
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    You're right, it's actually 20t-2t^2 and when it reaches a height of 50
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  4. #4
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    I am assuming t is time. As said max height is 50 at t=5. Due to symmetry of the curve it should be [5-a, 5+a].

    20t-2t^2=40 gives you the solution.
    I think it is: [5-sqrt(5), 5+sqrt(5)].

    -O

    P.S. I knew that plot of an object is a parabolla, f(x)=ax^2+bx+c, where x is the horizontal range. BUT I am not sure if it is still a parobolla when it is written with respect to time: f(t). Would that trajectory happen on the Jupiter?
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  5. #5
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    So is that -5 and 5? That's what I got originally, but I thought it didn't make sense to have the interval start at a negative number.
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  6. #6
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    No the solution is:

    <br />
5-\sqrt{5} \; \; and \; \; 5+\sqrt{5}<br />
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  7. #7
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    Thanks for your help! Did you use the quadratic formula to solve it? I used factoring and it seemed to work fine, so I wasn't sure where the square roots came from
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  8. #8
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    20t-2t^2=40
    -2t^2+20t-40=0
    t^2 -10t - 20=0
    disc.= 10^2 - 4*1*(-20)=20
    t1= (10-sqrt(20))/2
    t2= (10+sqrt(20))/2

    gives you the answer. You can factorize as well, but the above is easier.

    -O
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  9. #9
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    Quote Originally Posted by rebak View Post
    You're right, it's actually 20t-2t^2 and when it reaches a height of 50
    If I understand you correctly you have to solve

    20t-2t^2 \geq 50~\implies~2t^2-20t+50\leq 0~\implies~t^2-10t+25 \leq 0~\implies~(t-5)^2\leq 0

    The LHS of the inequality is a square which never will be negative (or smaller than zero - that's the same). At t = 5 the LHS will be zero.

    Thus the answer is: At t = 5 the ball reaches a height of 50'.
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  10. #10
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    Quote Originally Posted by oswaldo View Post
    No the solution is:

    <br />
5-\sqrt{5} \; \; and \; \; 5+\sqrt{5}<br />
    No, the question was "For what time interval is the ball higher than 40 feet?" The times the ball is at 40 feet are 5-\sqrt{5} and 5+\sqrt{5}. For any time between those two the ball is above 40 feet.
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  11. #11
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    Quote Originally Posted by earboth View Post
    If I understand you correctly you have to solve

    20t-2t^2 \geq 50~\implies~2t^2-20t+50\leq 0~\implies~t^2-10t+25 \leq 0~\implies~(t-5)^2\leq 0

    The LHS of the inequality is a square which never will be negative (or smaller than zero - that's the same). At t = 5 the LHS will be zero.

    Thus the answer is: At t = 5 the ball reaches a height of 50'.
    Earboth, I understood this to mean that with this new function the ball reaches a maximum height of 50 feet but that the problem was still to find the time interval when the ball was above 40 feet.
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