Question

An object is thrown upward from the top of a 112-foot building with an initial velocity of 96 feet per second. The height b of the object after t seconds is gubernatorial but the quadratic question The object will hit the ground when the time is ? seconds

Answer 1

We know that the height of the object is given by:

$h=-16t^2+96t+112$

To determine when the object will hit the ground we plug h=0 and solve the equation using the quadratic formula:

$\begin{gathered} -16t^2+96t+112=0 \\ t=\frac{-96\pm\sqrt[]{(96)^2-4(-16)(112)}}{2(-16)} \\ t=\frac{-96\pm\sqrt[]{16384}}{-32} \\ t=\frac{-96\pm128}{-32} \\ \text{then} \\ t=\frac{-96+128}{-32}=\frac{32}{-32}=-1 \\ \text{and} \\ t=\frac{-96-128}{-32}=\frac{-224}{-32}=7 \end{gathered}$

Since time has to be positve we conclude that the object will hit the ground when the time is 7 seconds.