An object is thrown off a 256-foot-tall building, and the distance of the object from the ground is measured every second. The function that models the height, h, of the object after t seconds is h(t) = –16t2 + 96t + 256. Determine the time when the object hits the ground

Let [tex]h(t) =-16\cdot t^{2}+96\cdot t + 256[/tex], where [tex]h[/tex] is the height of the object above the ground, measured in feet, and [tex]t[/tex] is time, measured in seconds. We notice that given expression is a second-order polynomial and hence, there are two roots, one of them corresponding to the instant when the object hits the ground. If [tex]h(t) = 0[/tex], we get the following expression:

[tex]-16\cdot t^{2}+96\cdot t +256 = 0[/tex] (Eq. 1)

Roots are found by applying the Quadratic Formula:

[tex]t_{1,2} = \frac{-96\pm\sqrt{96^{2}-4\cdot (-16)\cdot (256)}}{2\cdot (-16)}[/tex]

[tex]t_1 = 8\,s[/tex] or [tex]t_{2} = -2[/tex]

Only the first root offer a solution that is physically reasonable. Thus, the object hits the ground 8 seconds after launch.