Example 9 from Algebra and Trigonometry, 5.1 Quadratic Functions

Applying the Vertex and x-Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation $H (t) = - 16 t^{2} + 80 t + 40.$

ⓐWhen does the ball reach the maximum height?
ⓑWhat is the maximum height of the ball?
ⓒWhen does the ball hit the ground?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐThe ball reaches the maximum height at the vertex of the parabola.
  $\begin{matrix} h & = & - \frac{80}{2 (−16)} \\ = & \frac{80}{32} \\ = & \frac{5}{2} \\ = & 2.5 \end{matrix}$
  The ball reaches a maximum height after 2.5 seconds.
2. ⓑTo find the maximum height, find the $y -$ coordinate of the vertex of the parabola.
  $\begin{matrix} k & = & H (- \frac{b}{2 a}) \\ = & H (2.5) \\ = & −16 {(2.5)}^{2} + 80 (2.5) + 40 \\ = & 140 \end{matrix}$
  
  The ball reaches a maximum height of 140 feet.
3. ⓒTo find when the ball hits the ground, we need to determine when the height is zero, $H (t) = 0.$
  We use the quadratic formula.
  
  $\begin{matrix} t & = & \frac{−80 \pm \sqrt{80^{2} - 4 (−16) (40)}}{2 (−16)} \\ = & \frac{−80 \pm \sqrt{8960}}{−32} \end{matrix}$
  Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.
  $\begin{array}{l} \begin{array}{l} t = \frac{- 80 - \sqrt{8960}}{- 32} \approx 5.458 & or & t = \frac{- 80 + \sqrt{8960}}{- 32} \approx - 0.458 \end{array} \end{array}$
  The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16.
  
  Figure 16
  
  Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation $H (t) = −16 t^{2} + 96 t + 112.$

ⓐWhen does the rock reach the maximum height?
ⓑWhat is the maximum height of the rock?

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

ⓐWhen does the rock reach the maximum height?
ⓑWhat is the maximum height of the rock?

Solution

Your turn

Solution

Your turn