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Math Help - Solve the problem

  1. #1
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    Solve the problem

    A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been lanuched.

    My work (im lost)

    H = 48t - 16t^2

    Step 1 (t + 12) (t + 4t)

    Step 2 cancel out t left with (-12) (4t)

    Step 3 simplify -12/4 4t/4t

    Step 4 i am so lost your help is appreciated proper steps please.



    Thanks


    Mr.H
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Mr.Huge View Post
    A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been lanuched.

    My work (im lost)

    H = 48t - 16t^2

    Step 1 (t + 12) (t + 4t)

    Step 2 cancel out t left with (-12) (4t)

    Step 3 simplify -12/4 4t/4t

    Step 4 i am so lost your help is appreciated proper steps please.



    Thanks


    Mr.H
    What were you trying to do ?

    h(t)=48t-16t^2

    And you want the height when t=2.5

    h(2.5)=48 \cdot 2.5-16 \cdot (2.5)^2

    ..
    Last edited by Moo; June 25th 2008 at 02:27 PM. Reason: 16, not 18 ^^
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Moo View Post
    Hello,


    What were you trying to do ?

    h(t)=48t-18t^2

    And you want the height when t=2.5

    h(2.5)=48 \cdot 2.5-18 \cdot (2.5)^2

    ..
    \color{red}h(t)=48t-16t^2

    I know....same method! Hard to get a post in these days. I have to take what I can get.
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  4. #4
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    im still lost

    i still don't get how to work the problem. Can you be more detailed? Give me the run down of how to do this type of problem? I dont get it at all..sorry to sound so slow...this is all new to me.



    Mr.H
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  5. #5
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    Try these projectile word problems here....Quadratic Word Problems: Projectile Motion

    There are several examples.

    Good luck!
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