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27 January, 06:02

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h (t) = 52t - 16t^2. What is the maximum height that the ball will reach?

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  1. 27 January, 07:24
    0
    42.25 feet.

    Step-by-step explanation:

    The maximum height can be found by converting to vertex form:

    h (t) = 52t - 16t^2

    h (t) = - 16 (t^2 - 3.25t)

    h (t) = - 16 [ (t - 1.625) ^2 - 2.640625 ]

    = - 16 (t - 1.625 ^2) + 42.25

    Maximum height = 42.25 feet.

    Another method of solving this is by using Calculus:

    h (t) = 52t - 16t^2

    Finding the derivative:

    h' (t) = 52 - 32t

    This = zero for a maximum/minimum value.

    52 - 32t = 0

    t = 1.625 seconds at maximum height.

    It is a maximum because the path is a parabola which opens downwards. we know this because of the negative coefficient of x^2.

    Substituting in the original formula:h (t) = 52 (1.625) - 16 (1.625) ^2

    = 42.25 feet.
  2. 27 January, 07:31
    0
    42.25 feet

    Step-by-step explanation:

    The height function is a parabola. The maximum value of a negative parabola is at the vertex, which can be found with:

    x = - b/2a

    where a and b are the coefficients in y = ax² + bx + c.

    Here, we have y = - 16t² + 52t. So a = - 16 and b = 52. The vertex is at:

    t = - 52 / (2*-16)

    t = 13/8

    Evaluating the function:

    h (13/8) = - 16 (13/8) ² + 52 (13/8)

    h (13/8) = - 169/4 + 169/2

    h (13/8) = 169/4

    h (13/8) = 42.25
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