Question

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

Answer 1

The object hits the ground 8 seconds after launch.

Explanation:

Let
`h(t) =-16\cdot t^(2)+96\cdot t + 256`, where
`h` is the height of the object above the ground, measured in feet, and
`t` 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
`h(t) = 0`, we get the following expression:

`-16\cdot t^(2)+96\cdot t +256 = 0` (Eq. 1)

Roots are found by applying the Quadratic Formula:

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

`t_1 = 8\,s` or
`t_(2) = -2`

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