The equation S = -16t2 + 34t + 184 models the height of a ball that is thrown upward from the roof of a 184 foot building and falls to the street below. In this equation S is the height in feet of the ball above the ground and t is the time in seconds the ball has traveled. According to this model, how many seconds did it take the ball to reach a height of 91 feet? (round to 1 decimal place)

Answer:t=1.6Step-by-step explanation:If the equation [tex]S=-16t^2+34t+184[/tex] models the height of a ball above the ground, where t is time the ball travelled.If the height of the ball above the ground is S=91 instead of 184, then the time the ball should take to get to the ground comes from the expression above: [tex]S=-16t^2+34t+91[/tex] (because now we want to know how much time does it takes to reach the ground if it is thrown from 91 foot, not 184). Then, to know when the ball reaches the floor, we must equal the equation to zero [tex]-16t^2+34t+91=0[/tex] (because when the equation is zero, the height of the ball is zero, which means it is in the ground).To obtain the value of t in the expression [tex]-16t^2+34t+91=0[/tex] , we can apply the well known formula [tex]t=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex], where a is the coefficient that accompanies the quadratic term (in this case a=16), b is the coefficient that accompanies the linear term (b=34 in this case), and c is the constant coefficient (c=91).Because time is always possitive, we only retain the possitive value for t that solves the equation:[tex]\frac{-34(+-)\sqrt{34^2-4\times(-16)\times93} }{2\times16}[/tex]. [tex]t=1.55\simeq1.6[/tex]