a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.equation: y=-16x^2+153x+98

Answer:The rocket hits the ground at a time of 11.59 seconds.Step-by-step explanation:The height of the rocket, after x seconds, is given by the following equation:[tex]y = -16x^2 + 177x + 98[/tex]It hits the ground when [tex]y = 0[/tex], so we have to find x for which [tex]y = 0[/tex], which is a quadratic equation.Finding the roots of a quadratic equation:Given a second order polynomial expressed by the following equation:[tex]ax^{2} + bx + c, a\neq0[/tex].This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex][tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex][tex]\bigtriangleup = b^{2} - 4ac[/tex]In this question:[tex]y = -16x^2 + 177x + 98[/tex][tex]-16x^2 + 177x + 98 = 0[/tex]So[tex]a = -16, b = 177, c = 98[/tex][tex]\bigtriangleup = 177^{2} - 4(-16)(98) = 37601[/tex][tex]x_{1} = \frac{-177 + \sqrt{37601}}{2*(-16)} = -0.53[/tex][tex]x_{2} = \frac{-177 - \sqrt{37601}}{2*(-16)} = 11.59[/tex]Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.