Chapter 4, Problem 26CT

Solution

To find both the maximum altitude of the rocket and how long it takes to reach that altitude.

Maximum altitude of the rocket is $235.92\text{ m}$ and it takes $6.93\text{ seconds}$ to reach that altitude.

Given:

The function is given by:

  $h=68t-4.9t^2$

Formula used:

The axis of symmetry for the parabola with the equation of the form $h=a{t}^2+bt+c$ is given by:

  $t=-\frac{b}{2a}$

Calculation:

Comparing the given function with the parabola equation of the form $h=a{t}^2+bt+c$ , the parameters are:

  $\begin{array}{l}a=-4.9\\ b=68\\ c=0\end{array}$

Now, the axis of symmetry is given by:

  $\begin{array}{l}t=-\frac{b}{2a}\\ \Rightarrow t=-\left(\frac{\left(68\right)}{2\left(-4.9\right)}\right)\\ \Rightarrow t=\frac{68}{9.8}\\ \Rightarrow t=6.93\end{array}$

Hence, for the graph of given function, the axis of symmetry is $t=6.93$ . Hence, lightly sketch the line $t=6.93$ on the graph.

Now, the vertex of parabola is at $t=6.93$ . At $t=6.93$ , the $h-\text{coordinate}$ is

  $\begin{array}{l}$ h=68(6.93)-4.9{(6.93)}^2$\Rightarrow h=471.24-235.32\\ \Rightarrow h=235.92\end{array}$

Hence, plot the vertex $(6.93,235.92)$ on the graph. Since $c=0,$ the $y-\text{intercept}$ is $(0,0).$ The reflection of $(0,0)$ across $t=6.93$ is $(13.86,0).$ Hence, plot both the points on the graph.

Since $a<0,$ the parabola opens downward. Sketching a smooth curve through the points to draw the graph.

Graph:

  

Interpretation:

From the graph, the highest altitude is $h=235.92$ at $t=6.93$ . Hence, the maximum altitude of the rocket is $235.92\text{ m}$ and it takes $6.93\text{ seconds}$ to reach that altitude.