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# A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $s(t)=96 t-16 t^{2}.$ For what time t is the ball more than 128 feet above the ground?

Solution

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First, graph the function $f(t)=-16t^2+96t-128$

$\begin{array}{llll} \text{y-intercept: }&f(0)=-128 \; -\text{\textgreater} \fbox{(0,-128)}\\\\ \text{t-intercept: }&-16t^2+96t-128=0\\ &-16(t^2-6t+8)=0\\ &-16(t-4)(t-2)=0\\ &t-4=0,t-2=0\\ &t=4,\;t=2\\ &\fbox{(4,0)},\fbox{(2,0)}\\\\ \text{vertex: } &t=\dfrac{-b}{2a}=\dfrac{-96}{2(-16)}=\dfrac{-96}{-32}=3\\\\ f(3)&t=-16(3)^2+96(3)-128\\&=-144+288-128=16\\ &\fbox{(3,16)}\\ \\ \end{array}$

Since the distance from the ground is $s(t)$, you want to solve the equation when the distance from the ground $\geq$ 128.

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