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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)=96t16t2.s(t)=96 t-16 t^{2}. For what time t is the ball more than 128 feet above the ground?


Answered 2 years ago
Answered 2 years ago
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First, graph the function f(t)=16t2+96t128f(t)=-16t^2+96t-128

y-intercept: f(0)=128  >(0,-128)t-intercept: 16t2+96t128=016(t26t+8)=016(t4)(t2)=0t4=0,t2=0t=4,  t=2(4,0),(2,0)vertex: t=b2a=962(16)=9632=3f(3)t=16(3)2+96(3)128=144+288128=16(3,16)\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)s(t), you want to solve the equation when the distance from the ground \geq 128.

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