The height h (in meters) of a cannonball t seconds after it is fired into the air is given by

a) What is the initial height of the cannonball? What feature does this correspond to on the graph?

b) Find the height of the cannonball after 1 second.

c) What is the maximum height of the cannonball? When does it reach the maximum height?

d)How long does it take for the cannonball to fall back to earth?

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  1. a) Hey, you know what h is when t = zero!

    b ) h = -4(1) + 16(1) + 9
    = 21

    c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola
    t^2 - 4 t = -h/4 + 9/4

    t^2 - 4 t + 4 = -h/4 + 9/4 + 16/4

    (t - 2)^2 = -(1/4) (h - 25)
    max height = 25 at t = 2

    d ) 0 =-4t^2+16t+9
    t^2 - 4 t - 9/4 = 0

    t = [ 4 +/- sqrt(16 -81/4) ] /2
    no real solution
    The reason is that your original equation is in error
    It should be assuming feet and seconds and g = -32 ft/s^2
    h = (1/2) g t^2 + Vi t + Hi
    h = -16 t^2 + Vi t + Hi
    I suspect a typo

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  2. I agree with Damon that your gravitational constant is off.
    Since you are using metric , your equation should be

    h = -4.905t^2 + 16t + 9
    let's ballpark it to

    h = -5t^2 + 16 + 9

    a) .....

    b) sub in t = 1 , h = -5+16+9 = 20 m

    c) for t of the vertex ...
    t =-b/(2a) = -16/-10 = 1.6
    h = -5(1.6^2) + 16(1.6) = 9 = 21.8

    Maximimum height of 21.8 m after 1.6 seconds

    d) -5t^2 + 16t + 9 = 0
    5t^2 - 16t - 9 = 0
    t = (16 ± √436)/10
    = 3.69 or a negative

    it will hit the ground after 3.69 seconds

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  3. d) use quadratic formula
    -b +- square root of b^2 - 4 times a times c devided by 2a
    = -16+-square root 400 devided by -8
    = 4.5 secounds.

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