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Thread: 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,
    $\displaystyle 80t-16t^2=0$
    $\displaystyle 16t(5-t)=0$
    $\displaystyle t=0,5$

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

    Thus, the time intervals are:
    $\displaystyle 0\leq t\leq 1 \mbox{ and } 4\leq t \leq 5$
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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 $\displaystyle s$ (in feet) of the ball after $\displaystyle t$ seconds is: $\displaystyle s \:= \:80t - 16t^2$
    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: .$\displaystyle 80t - 16t^2 \:=\:60\quad\Rightarrow\quad t^2 - 5t + 4 \:=\:0$
    . . $\displaystyle (t - 1)(t - 4)\:=\:0\quad\Rightarrow\quad t \:=\:1,\,4$

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


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

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

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

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

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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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