College Algebra, Chapter 1, 1.3, Section 1.3, Problem 96

Suppose that an object is dropped from a height $96 ft$ tall above the ground. Then its height after $t$ seconds is given by $h = -16t^2 + 96$, where $h$ is measured in feet.

a.) How long will it take to fall half the distance to ground level?


$\begin{equation} \begin{aligned} h =& -16t^2 + 96 && \text{Model} \\ \\ \frac{96}{2} =& -16t^2 + 96 && h = \frac{96}{2} \text{ half the distance to ground level} \\ \\ 16t^2 =& 96 - \frac{96}{2} && \text{Add } 16t^2 \text{ and subtract } \frac{96}{2} \\ \\ t^2 =& \frac{48}{16} = 3 && \text{Solve for } t \\ \\ t =& \pm \sqrt{3} && \text{Take the square root} \\ \\ t =& \sqrt{3} \text{ and } t = \sqrt{3} && \text{Choose } t > 0 \\ \\ t =& \sqrt{3} \text{ seconds} && \end{aligned} \end{equation}$


b.) How long will it take to fall to ground level?


$\begin{equation} \begin{aligned} h =& -16t^2 + 96h = -16t^2 + 96 && \text{Model} \\ \\ 0 =& -16t^2 + 96 && h = 0 \text{ ground level} \\ \\ 16t^2 =& 96 && \text{Add } 16t^2 \\ \\ t^2 =& \frac{96}{16} && \text{Divide } 16 \\ \\ t^2 =& 6 && \text{Simplify} \\ \\ t =& \pm \sqrt{6} && \text{Take the square root} \\ \\ t =& \sqrt{6} \text{ and } t = - \sqrt{6} && \text{Choose } t > 0 \end{aligned} \end{equation}$