Chapter 6.1, Problem 58PPSE

Solution

To choose the correct option.

Option $\left(C\right)$

Given:

The given equation as:

  $h\left(t\right)=-5t^2+70t$

where

  $\begin{array}{l}h=\text{ height in feet}\\ t=\text{ time in seconds}\end{array}$

Calculation:

If a quadratic function is given in the standard form $f\left(x\right)=a{x}^2+bx+c,$ then the point $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ is the vertex of the parabola.

Compare the given equation with $f\left(t\right)=a{t}^2+bt+c,$ the parameters are:

  $\begin{array}{l}a=-5\\ b=70\\ c=0\end{array}$

Now, finding the $x$ -coordinate of vertex as:

  $\begin{array}{l}\frac{-b}{2a}=\frac{-70}{2\left(-5\right)}\\ \Rightarrow \frac{-b}{2a}=\frac{-70}{-10}\\ \Rightarrow \frac{-b}{2a}=7\end{array}$

Now, substitute the $\frac{-b}{2a}$ for $t$ in the equation to finding the $y$ -coordinate of the vertex as:

  $\begin{array}{l}h\left(\frac{-b}{2a}\right)=-5{\left(\frac{-b}{2a}\right)}^2+70\left(\frac{-b}{2a}\right)\\ \Rightarrow h\left(7\right)=-5{\left(7\right)}^2+70\left(7\right)\\ \Rightarrow h\left(7\right)=-5\times 49+70\times 7\\ \Rightarrow h\left(7\right)=-245+490\\ \Rightarrow h\left(7\right)=245\end{array}$

Hence, the vertex of the parabola $h\left(t\right)=-5t^2+70t$ is

  $\left(\frac{-b}{2a},h\left(\frac{-b}{2a}\right)\right)=\left(7,245\right)$

Now, the parabola $h\left(t\right)=-5t^2+70t$ is a downward parabola, as its leading coefficient is negative. Hence, the vertex of the parabola would be the highest point. It means the rocket reaches the maximum height at $7\text{ seconds}\text{.}$ As the time between ignitions is $2\text{ seconds}$ and the second ignition occurs at $7\text{ seconds}\text{.}$ The first ignition must occur at

  $7-2=5\text{ seconds}$

Hence, the timer of the first ignitions should be set to $5\text{ seconds}$ after the shot. So, Option $\left(C\right)$ is correct answer.