Math 10

Hello,
Is there easiest and simplest way to do this question?

"A baseball is thrown from the top of a building and falls to the ground below. The height of the baseball above the ground is approximated by the relation h = -5t2 + 10t + 15, where h is the height above the ground in metres and "t" is the elapsed time in seconds. Determine the maximum height that is reached by the ball."

Thank you very much!

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  1. that is a parabola (quadratic)

    5 t^2 - 10 t - 15 = -h

    find the vertex by completing the square

    t^2 - 2 t - 7.5 = -(1/5)(h)

    t^2 - 2 t = -h/5 + 7.5

    t^2 - 2 t + 1 = -h/5 + 8.5

    (t-1)^2 = -(1/5) (h - 42.5)

    so max height at one second of 42.5 meters

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  2. sure, all you have to do is to find the vertex of this "downwards" parabola.

    I will assume that you do not know Calculus , so
    (easiest way):

    the x of the vertex for y = ax^2 + bx + c
    is -b/(2a)
    so for your equation, the
    t of the vertex is -10/-10 = 1
    sub into the equation ....
    h = -5(1^2) + 10(1) + 15
    = -5 + 10 + 15 = 20
    vertex is (1,20)

    so the ball reaches a max of 20 m after 1 second

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  3. t^2 - 2 t - 3 = -(1/5)(h)

    t^2 - 2 t = -h/5 + 3

    t^2 - 2 t + 1 = -h/5 + 4

    (t-1)^2 = -(1/5) (h - 20)

    so max height at one second of 20 meters

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  4. Thanks alot guys! Both of your answers were very helpful.

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