Chapter SH, Problem 4.5EP

To determine

Find the time takenby the rocket in air.

Answer to Problem 4.5EP

The rocket is in the air are $4$ seconds.

Explanation of Solution

Given:

The given modelis $h=-16t^2+64t$ , where $h$ is the height of the rocket and $t$ is the time after rocket is launched.

Calculation:

Draw a graph for the given model $h=-16t^2+64t$ .

Equation of the axis of symmetry.

  $\begin{array}{l}x=-\frac{b}{2a}\\ \Rightarrow t=-\frac{64}{2\cdot \left(-16\right)}\\ \Rightarrow t=2\end{array}$

The axis of symmetry intersects a parabola at only one point, called the vertex.

The x-coordinate of the vertex $=-\frac{b}{2a}$

  $=-\frac{64}{2\cdot \left(-16\right)}$

  $=2$

Substitute the value of $t=2$ in the given function.

  $\begin{array}{l}h=-16{\left(2\right)}^2+64\left(2\right)\\ =-64+128\\ =64\end{array}$

Vertex $=\left(2,64\right)$

Draw a table for the values of $h$ with respect to $t$ .

$t$	$h=-16t^2+64t$	$\left(t,h\right)$
$\begin{array}{l}0\\ 1\\ 2\\ 3\\ 4\end{array}$	$\begin{array}{l}0\\ 48\\ 64\\ 48\\ 0\end{array}$	$\begin{array}{l}\left(0,0\right)\\ \left(1,48\right)\\ \left(2,64\right)\\ \left(3,48\right)\\ \left(4,0\right)\end{array}$

The graph of the function is given below.

  

The solution of a quadratic equation are called the roots of the equation. One method for finding the roots of a quadratic is to find the zeros of the related quadratic function.

The zeros of the function are the x-intercepts of the graph. The x-intercepts are the point(s) at which the parabola intersects the x-axis.

In the given graph, the x-intercepts are $0$ and $4$ .

Hence the rocket is in the air are $4$ seconds.