Chapter 6.8, Problem 17STP

(a)

To determine

To find:maximum height of the rocket and round to nearest tenth of a meter.

Answer to Problem 17STP

The maximum height is $160\text{m}$ .

Explanation of Solution

Given:

  $h=-4.9t^2+56t$ .

Concept used:

If $f^{″}\left(x\right)>0$ then the curve will open- upward and the point is Minima.

Similarly, if $f^{″}\left(x\right)<0$ then the curve will open downward and the point is Maxima.

Calculation:

According to the given:

The equation of rocket launching is given by:

  $h=-4.9t^2+56t$ .

Where $h$ is height of the rocket meter $t$ is time in second after launch of rocket.

  $\begin{array}{l}h=-4.9t^2+56t.\\ \frac{d\left(h\right)}{dt}=-9.8t+56.\\ \frac{d^2\left(h\right)}{d{t}^2}=-\mathrm{9.8.}\end{array}$

According to the maxima minima formula:

If $f^{″}\left(x\right)>0$ then the curve will open- upward and the point is Minima.

Similarly, if $f^{″}\left(x\right)<0$ then the curve will open downward and the point is Maxima.

So, here the point will be Maxima and the point of Maxima can be calculated as:

  $\begin{array}{l}{f}^{\prime }\left(h\right)=\frac{dh}{dt}=0.\\ \Rightarrow -9.8t+56=0.\\ \Rightarrow -9.8t=-56.\\ t=5.7143\approx 5.7\end{array}$

At $t=5.7143\sec$ the rocket will reach the maximum height and the maximum height can be calculate as:

  $\begin{array}{l}h\left(5.7143\right)=-4.9{\left(5.7143\right)}^2+56\left(5.7143\right).\\ \Rightarrow -4.9{\left(5.7143\right)}^2+56\left(5.7143\right).\\ h=159.9999\approx 160.\end{array}$

Hence, the maximum height is $160\text{m}$ .

(b)

To determine

To find: The Domain of the function.

Answer to Problem 17STP

The maximum time is $t=5.7$ .

Explanation of Solution

Given:

  $h=-4.9t^2+56t$ .

Concept used:

the point will be Maxima and the point of Maxima can be calculated as:

  $f^{\prime }\left(h\right)=\frac{dh}{dt}=0$ .

Calculation:

The time taken to reach the maxima height is:

  $\begin{array}{l}{f}^{\prime }\left(h\right)=\frac{dh}{dt}=0.\\ \Rightarrow -9.8t+56=0.\\ \Rightarrow -9.8t=-56.\\ t=5.7143\approx 5.7\end{array}$

Hence, the maximum time is $t=5.7$ .