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Math Help - Quadratics

  1. #1
    Super Member sakonpure6's Avatar
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    Quadratics

    Can some one please verify my work? Thank you!

    Tom throws a baseball so that the height of the ball is given by h(t)= -5t^2 + 20t + 1 where h is the height reached in meters and t is the time taken in seconds.

    a) If the ball is caught at the same height it is thrown, how long was it in the air?

    What i did:

    set h(t) = 1 because that is the height when we through the ball, and solve.

    1= -5t^2 +20t + 1
    0= -5t^2 + 20t
    0= -5t(t - 4)

    There fore t = 4 & 0, and the ball was in the air for 4 seconds.

    b) IF the ball is caught at the same height at which it is thrown, what is the maximum height reached by the ball?

    What I did:

    We have two symmetrical pairs ( 0,1) and (4,1).

    So the AOS:
    x= (4+0)/2
    x=2

    Sub x=2 in the equation to solve for max. height!

    h(2)= -5(2)^2+20(2)+1
    =21
    There fore the max height is 21.
    Last edited by sakonpure6; January 26th 2013 at 02:40 PM.
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    Re: Quadratics

    Quote Originally Posted by sakonpure6 View Post
    a) If the ball is caught at the same time it is thrown, how long was it in the air?
    How can you catch a ball at the same time it is thrown?
    I am confused by the wording of the question, can you clarify please?

    I looked at your work in part (a), it seems logical. In part (b), I don't get what you mean by "AOS". Is this question from a physics book?
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  3. #3
    Super Member sakonpure6's Avatar
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    Re: Quadratics

    Quote Originally Posted by obitus View Post
    How can you catch a ball at the same time it is thrown?
    I am confused by the wording of the question, can you clarify please?

    I looked at your work in part (a), it seems logical. In part (b), I don't get what you mean by "AOS". Is this question from a physics book?
    Ops sorry, it should be "... at the same height it is thrown"

    and for part b) AOS stands for Axis of symmetry which is also the x value of the vertex.
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    Re: Quadratics

    Quote Originally Posted by sakonpure6 View Post
    b) IF the ball is caught at the same height at which it is thrown, what is the maximum height reached by the ball?
    In part (a) you set the initial height to 1, why did you not set the initial height to 0? Where 0 represents the position from the ball leaving Tom's hand. As I stated earlier, your mathematical work is sound.
    <br />
1 = (-5)(t)^{2} + (20)(t) + 1<br />

    You start off by subtracting 1 on both sides, to yield:

    <br />
0 = (-5)(t)^{2} + (20)(t)<br />

    Then you factored out -5t, to yield:

    <br />
0 = (-5)(t) \times (t - 4)<br />

    Then you solved for t and got:
    <br />
t = 0 ,4<br />

    All that makes sense to me, what I don't get is the logic of setting the height to 1. For part (b), if you know the time of ball in the air then could you not calculate the height for half the total time?
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  5. #5
    Super Member sakonpure6's Avatar
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    Re: Quadratics

    Quote Originally Posted by obitus View Post
    In part (a) you set the initial height to 1, why did you not set the initial height to 0? Where 0 represents the position from the ball leaving Tom's hand. As I stated earlier, your mathematical work is sound.
    <br />
1 = (-5)(t)^{2} + (20)(t) + 1<br />

    You start off by subtracting 1 on both sides, to yield:

    <br />
0 = (-5)(t)^{2} + (20)(t)<br />

    Then you factored out -5t, to yield:

    <br />
0 = (-5)(t) \times (t - 4)<br />

    Then you solved for t and got:
    <br />
t = 0 ,4<br />

    All that makes sense to me, what I don't get is the logic of setting the height to 1. For part (b), if you know the time of ball in the air then could you not calculate the height for half the total time?
    The height can not be 0 because if you set time = 0 (which is when the ball is thrown) and solve the equation, you get an initial height of 1!

    how about for part b) do you agree with what I did?
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    Re: Quadratics

    Quote Originally Posted by obitus View Post
    In part (a) you set the initial height to 1, why did you not set the initial height to 0? Where 0 represents the position from the ball leaving Tom's hand.
    What are you taking as the time the ball was initially thrown? Why are you assuming the ball was thrown from height 0? The word of the problem, "t is the time taken in seconds", implies that t measures the time it takes the ball to go from one person to the other- that is the ball was thrown when t= 0 so its height at the at time is -5(0^2)+ 20(0)+ 1= 1 meter as sakonpure6 used.

    As I stated earlier, your mathematical work is sound.
    <br />
1 = (-5)(t)^{2} + (20)(t) + 1<br />

    You start off by subtracting 1 on both sides, to yield:

    <br />
0 = (-5)(t)^{2} + (20)(t)<br />

    Then you factored out -5t, to yield:

    <br />
0 = (-5)(t) \times (t - 4)<br />

    Then you solved for t and got:
    <br />
t = 0 ,4<br />

    All that makes sense to me, what I don't get is the logic of setting the height to 1. For part (b), if you know the time of ball in the air then could you not calculate the height for half the total time?
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  7. #7
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    Re: Quadratics

    Quote Originally Posted by sakonpure6 View Post
    Can some one please verify my work? Thank you!

    Tom throws a baseball so that the height of the ball is given by h(t)= -5t^2 + 20t + 1 where h is the height reached in meters and t is the time taken in seconds.

    a) If the ball is caught at the same height it is thrown, how long was it in the air?

    What i did:

    set h(t) = 1 because that is the height when we through the ball, and solve.

    1= -5t^2 +20t + 1
    0= -5t^2 + 20t
    0= -5t(t - 4)

    There fore t = 4 & 0, and the ball was in the air for 4 seconds.

    b) IF the ball is caught at the same height at which it is thrown, what is the maximum height reached by the ball?

    What I did:

    We have two symmetrical pairs ( 0,1) and (4,1).
    Yes, this is correct. A parabola is symmetric about its vertex, which, in this case, is the greatest height of the ball. Another way to do this is to "complete the square". We can write [tex]h= -5t^2+ 20t+ 1= -5(t^2- 4t)+ 1= -5(t^2- 4t+ 4- 4)+ 1= -5(t^2- 4t+ 4)+ 20+ 1= -5(t^2- 4t+ 4)+ 21. The reason I added and subtracted 4, and then took the "-4" outside the parentheses, was to get a "perfect square" inside the parentheses: t^2- 4t+ 4= (t- 2)^2.

    That gives h= -5(t- 2)^2+ 21. Since a square is never negative, -5(t- 2)^2 is never positive. At t= 2 (yes, halfway between 0 and 4), h= 21 meters. For any other t we have 21 minus something so that 21 meter is, in fact, the greatest height.

    So the AOS:
    x= (4+0)/2
    x=2

    Sub x=2 in the equation to solve for max. height!

    h(2)= -5(2)^2+20(2)+1
    =21
    There fore the max height is 21.[/QUOTE]
    Thanks from sakonpure6
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    Re: Quadratics

    Oh that makes sense, I was thinking a ball thrown straight up into the air and free falling back down. Thanks for clearing that up.
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  9. #9
    Super Member sakonpure6's Avatar
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    Re: Quadratics

    Thank for all the help!
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