A projectile is launched upward with a velocity of 288 feet per second from the top of a 65 foot structure . What is the maximum height atta

Question

A projectile is launched upward with a velocity of 288 feet per second from the top of a 65 foot structure . What is the maximum height attained by the projectile ?

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    0
    2021-11-23T20:46:06+00:00

    d = vt +  \frac{1}{2} g {t}^{2}  \\ 0 = 288t  -  5 {t}^{2}  \\ t = 0 \\ or \\ t =  \frac{ 288}{5}  \\  \frac{t}{2}  =   \frac{ 288}{10}  = 2.88 \: sec \\ d = 288(2.88) + 5 ({2.88}^{2} ) \\ d = 829.44 + 41.472 = 870.912 \\ h = 870.912 + 65 = 935.912 \: ft
    :3

    0
    2021-11-23T20:46:25+00:00

    Answer:

    The maximum height attained by the projectile is 412.94 meters = 1359.79 feet.

    Step-by-step explanation:

    The height formula has the following formula:

    h = h_{0} + v_{0}t - \frac{gt^{2}}{2}

    In which h_{0} is the initial height, v_{0} is the initial speed and g = 9.8 m/s² is the gravity.

    In this problem, we have that:

    We are working with the gravity in meters, so we must convert the feet measures to meters.

    Each feet has 0.3048 meters.

    So 288 feet per second = 87.78 meters per second.

    65 foot = 19.81 meters.

    This means that:

    h_{0} = 19.81, v_{0} = 87.78

    So

    h(t) = 19.81 + 87.78t - 4.9t^{2}

    The maximum height is attained at the moment of time in which the velocity is 0. The velocity is the derivative of the height. So:

    v(t) = h'(t) = -9.8t + 87.78

    v(t) = 0

    9.8t = 87.78

    t = 8.96

    The maximum height is attained at 8.96s. This height is

    h(t) = 19.81 + 87.78t - 4.9t^{2}

    h(8.96) = 19.81 + 87.78*8.96 - 4.9*(8.96)^{2} = 412.94

    The maximum height attained by the projectile is 412.94 meters = 1359.79 feet.

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