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Math Help - Ball Above Ground

  1. #1
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    Ball Above Ground

    A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 80t - 16t^2. For what time interval is the ball less than 64 feet above the ground?
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  2. #2
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    Quote Originally Posted by symmetry View Post
    A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 80t - 16t^2. For what time interval is the ball less than 64 feet above the ground?
    If returns back down at,
    80t-16t^2=0
    16t(5-t)=0
    t=0,5

    We need for 0\leq t\leq 5
    To satisfy,
    80t-16t^2 \leq 64
    Thus,
    16t^2-80t + 64 \geq 0
    t^2 - 5t + 4 \geq 0
    (t-1)(t-4) \geq 0
    We have two cases,
    1)Both positive.
    t-1>0
    t-4>0
    Thus,
    t>4
    2)Both negative,
    t-1<0
    t-4<0
    Thus,
    t<1.

    Thus, the time intervals are:
    0\leq t\leq 1 \mbox{ and } 4\leq t \leq 5
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  3. #3
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    Hello, symmetry!

    I'll baby-step through the explanation . . .


    A ball is thrown vertically upward with an initial velocity of 80 ft/sec.
    The height s (in feet) of the ball after t seconds is: s \:= \:80t - 16t^2
    For what time interval is the ball less than 64 feet above the ground?

    When is the height of the ball exactly 64 feet?

    We have: . 80t - 16t^2 \:=\:60\quad\Rightarrow\quad t^2 - 5t + 4 \:=\:0
    . . (t - 1)(t - 4)\:=\:0\quad\Rightarrow\quad t \:=\:1,\,4

    So the ball is takes 1 second to reach a height of 64 feet.
    . . The time interval is: . \boxed{[0,\,1)}


    Then it goes beyond 64 feet to its maximum height, then falls back down,
    . . passing the 64-foot mark at t = 4.
    So it is again below 64 feet from t = 4 until it hits the ground.
    . . Uh-oh! .When does it hit the ground?

    The ball is on the ground when s = 0.
    So we have: . 80t - 16t^2\:=\:0
    . . Then: . 16t(5 - t)\:=\:0\quad\Rightarrow\quad t \:=\:0,\,5

    So the ball is on the ground when t = 0 (of course!)
    . . and again when t = 5.

    So it is below 64 feet for last second: . \boxed{(4,\,5]}

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  4. #4
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    ok

    I thank you BOTH for your replies and guidance through the question.
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