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22 July, 08:44

An object is dropped from 43 feet below the tip of the pinnacle atop a 527 -ft tall building. The height h of the object after t seconds is given by the equation h equals negative 16 t squared plus 484. Find how many seconds pass before the object reaches the ground.

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  1. 22 July, 08:50
    0
    5.5 seconds

    Step-by-step explanation:

    The function that gives the height of the object after t seconds falling is:

    h = - 16t^2 + 484.

    To find the time when the object reaches the ground, we just need to use the value of h = 0 in the equation, and then find the value of t:

    0 = - 16t^2 + 484.

    16t^2 = 484

    t^2 = 30.25

    t = 5.5 seconds

    So the object will reach the ground after 5.5 seconds
  2. 22 July, 08:50
    0
    5.5 sec

    Step-by-step explanation:

    The height h of the object after t seconds is given by the equation

    h = - 16 t² + 484

    The height would be zero when the object hits the ground. therefore equating the above equation to zero

    0 = - 16 t² + 484 (solving for t)

    16 t² = 484

    t² = 484/16

    t² = 30.25

    taking square root on B. S

    t = 5.5 sec
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