An object is dropped from 24 feet below the tip of the pinnacle atop a 1468-ft tall building. The height h of the object after t seconds is

Question

An object is dropped from 24 feet below the tip of the pinnacle atop a 1468-ft tall building. The height h of the object after t seconds is given by the equatior h= – 16t2 + 1444. Find how many seconds pass before the object reaches the ground. seconds pass before the object reaches the ground. (Type an integer or a decimal.)​

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Hải Đăng 1 week 2021-07-22T09:23:28+00:00 1 Answers 1 views 0

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    2021-07-22T09:25:09+00:00

    Answer:

    9.5 seconds pass before the object reaches the ground.

    Step-by-step explanation:

    Height of the ball:

    The height of the ball after t seconds is given by the following equation:

    h(t) = -16t^2 + 1444

    Find how many seconds pass before the object reaches the ground.

    This is t for which h(t) = 0. So

    h(t) = -16t^2 + 1444

    -16t^2 + 1444 = 0

    16t^2 = 1444

    t^2 = \frac{1444}{16}

    t^2 = 90.25

    t = \pm \sqrt{90.25}

    Since it is time, we only take the positive value.

    t = 9.5

    9.5 seconds pass before the object reaches the ground.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )