Chapter 1.7, Problem 63PS

Solution

To find: the volleyball’s maximum height.

The maximum height of the volleyball is $40$ feet.

Given:

The height of a volleyball $t$ seconds after it is hit is modeled by $h=-16t^2+48t+4$ .

Concept used:

The vertex form of a quadratic function is $y=a{\left(x-h\right)}^2+k$ , where $\left(h,k\right)$ is the vertex of the function’s graph.

Calculation:

Consider, $h=-16t^2+48t+4$ .

Taking factor $-16$ from first two terms, it follows that $h=-16\left(t^2-3t\right)+4$ .

Add $-16{\left(\frac{3}{2}\right)}^2=-36$ to each side, it follows that

  $\begin{array}{c}h-16\left(\frac{9}{4}\right)=-16\left(t^2-3t+\frac{9}{4}\right)+4\\ h-36=-16{\left(t-\frac{3}{2}\right)}^2+4\\ h=-16{\left(t-\frac{3}{2}\right)}^2+40\end{array}$

The vertex is $\left(\frac{3}{2},40\right)$ .

So, the maximum height of the volleyball is $40$ feet.