Chapter 2.2, Problem 14Q

To determine

To find : The height travelled by flare and compare which one will stay longer in the air.

Answer to Problem 14Q

First flare travels higher than second flare. First flare will remain in the air longer.

Explanation of Solution

Given information : The height of the flare above the water is given by $f\left(t\right)=-16t\left(t-8\right)$ and the path of the second flare is given by the graph

  

From the given graph, height of the second flare is $\text{196 inches}$ . To find the height of the first flare, we need to find the vertex of $f\left(t\right)=-16t\left(t-8\right)$ .

Therefore, we need to write the $f\left(t\right)=-16t\left(t-8\right)$ as standard form as ${\left(x-\alpha \right)}^2=-a\left(y-\beta \right)$ where $\left(\alpha ,\beta \right)$ is the vertex.

  $\begin{array}{l}y=-16t\left(t-8\right)\\ -\frac{y}{16}=t^2-8t+16-16\\ -\frac{y}{16}+16={\left(t-4\right)}^2\\ {\left(t-4\right)}^2=-\frac{1}{16}\left(y-256\right)\end{array}$ .

Thus, $\left(\alpha ,\beta \right)=\left(4,256\right)$ is the vertex of $f\left(t\right)=-16t\left(t-8\right)$ . Therefore, height of the first flare is $256\text{ inches}$ which is greater than that of second flare.

Hence, first flare travels higher than second flare. Therefore, first flare will remain in the air longer.