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8 December, 15:34

A ball is thrown vertically upward with an initial velocity of 32 feet per second. The distance s (feet) of the ball from the ground after t seconds is s equals 32t-16t^2. At what time t will the ball strike the ground? For what time t is the ball more than 12 feet above the ground

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Answers (2)
  1. 8 December, 17:01
    0
    a) t = 2 sec

    b) 0.5s < t < 1.5s

    Explanation:

    Given dа ta:

    a) set s=given equation to zero and solve for t:

    s = 32t - 16t^2

    0 = 32t-16t^2

    0 = t (32-16t)

    t = 0

    32 = 16 t

    t = 2 seconds

    Answer: The ball strikes the ground again at t = 2 seconds

    b) set given equation greater than 12 and solve for t:

    s = 32 t - 16t^2

    12 < 32t-16t^2

    12 < 32t-16t^2

    t^2 - 2t + 0.75 < 0

    0.5s < t < 1.5s

    Answer: The ball is above 12 feet from the ground in the time interval of 0.5s < t < 1.5s
  2. 8 December, 17:50
    0
    a) t = 2 sec

    b) t = 0.5 sec

    Explanation:

    Given dа ta:

    a) set s=given equation to zero and solve for t:

    s = 32t - 16t^2

    0 = 32t-16t^2

    32t = 16t^2

    32 = 16 t

    t = 2 seconds

    b) set given equation equal to 12 and solve for t:

    s = 32 t - 16t^2

    12 = 32t-16t^2

    12 = 32t-16t^2

    t^2 - 2t + 0.75 = 0

    t = 1.5, 0.5

    hence t = 0.5 sec is right answer
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