The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of –30 ft/sec is modeled by the equation h(t) = –16t^2 – 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air? t = 2 –2 < t < 0 0 < t < 2 t > 2

Answer:Option 3.Step-by-step explanation:The given function is[tex]h(t)=-16t^2-30t+124[/tex]where, h(t) is the height of a small rock falling from the top of a 124-ft-tall building and t is the time in seconds.It is a downward parabola.Equate the function equal to 0, to find the time at which the rock touch the ground.[tex]-16t^2-30t+124=0[/tex]If [tex]ax^2+bx+c=0[/tex], the according to the quadratic formula [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]Using quadratic formula, we get[tex]t=\dfrac{-(-30)\pm \sqrt{(-30)^2-4(124)(-16)}}{2(-16)}[/tex][tex]t=\dfrac{30+\sqrt{(-30)^2-4(124)(-16)}}{2(-16)},\dfrac{30-\sqrt{(-30)^2-4(124)(-16)}}{2(-16)}[/tex][tex]t=-3.875, 2[/tex]Time cannot be negative. So, the rock remain in the air in the interval 0 < t < 2.Therefore, the correct option is 3.