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Math Help - finding height reached by cannon shell

  1. #1
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    finding height reached by cannon shell

    a cannon fires a shell that rise (h) cm after t seconds where
    h(t)=800t-16t^2

    how would i find the maximium height (k) reached by the shell and how would its graph look. for h=800t -16t^2
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  2. #2
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    You should be able to see that the graph is a parabola.

    Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.
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    am tryin to determine the constants A and B but am having no luck
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    Quote Originally Posted by HallsofIvy View Post
    You should be able to see that the graph is a parabola.

    Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.
    i got -16(t-25)^2+1000

    so i can assume that the maximium point is (25, 1000) is that correct?
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    Quote Originally Posted by HallsofIvy View Post
    You should be able to see that the graph is a parabola.

    Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.
    i got -16(t-25)^2+1000

    so i can assume that the maximium point is (25, 1000) is that correct?
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  6. #6
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    16*625= 100^2= 10000, not 1000.
    Last edited by HallsofIvy; April 11th 2010 at 05:55 AM.
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    Quote Originally Posted by sigma1 View Post
    a cannon fires a shell that rise (h) cm after t seconds where
    h(t)=800t-16t^2

    how would i find the maximium height (k) reached by the shell and how would its graph look. for h=800t -16t^2
    h(t) = 800t - 16t^2

    h(t) = 16t(50 - t)

    0 = 16t(50-t)

    note that the h = 0 at t = 0 and t = 50

    due to the parabola's symmetry, the shell will be at its highest point at t = 25, midtime between the zeros.

    h(25) = 16(25)[50-25] = 16 \cdot 25^2 = 10000
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    ooh thank you i now see that.. so i can say then that the maximium height reached is 10000 as opposed to stating the coordinates of (25,10000),

    i have also sketched this graph is it correct?
    Attached Thumbnails Attached Thumbnails finding height reached by cannon shell-untitled3.bmp  
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  9. #9
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    Quote Originally Posted by sigma1 View Post
    a cannon fires a shell that rise (h) cm after t seconds where
    h(t)=800t-16t^2

    how would i find the maximium height (k) reached by the shell and how would its graph look. for h=800t -16t^2
    x = ut+\frac{1}{2}at^2

    \frac{dx}{dt}= v = u+at

    so we have x=800t-16t^2
    <br />
\frac{dx}{dt}= v = 800-32t

    when v=0 it reaches max. height because when v is negative, it will start falling.

    so \rightarrow 800-32t=0

    rearrange to find t which reaches max height.

    which comes to 25s.

    then sub it into x=800t-16t^2, to find max height.

    which comes to 10000m.
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  10. #10
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    Quote Originally Posted by sigma1 View Post
    ooh thank you i now see that.. so i can say then that the maximium height reached is 10000 as opposed to stating the coordinates of (25,10000),

    i have also sketched this graph is it correct?
    Yes, that is correct.
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that is correct.
    thanks for that. but how would these transformation make the graph. look .could you show me?

    |h(t)|

    h(t+30)
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