Chapter 9.1, Problem 46PPE

Solution

(a)

To graph: The function $h=-16t^2+72$ .

Given information:

The given function $h=-16t^2+72$ .

Calculation:

Consider the given function $h=-16t^2+72$ .

Now, draw the graph,

  

Interpretation:

The domain of the graph is 72 and the range is $\left(-2.05,2.05\right)$ .

(b)

To calculate: The distance from 0 to 1 second.

The required height is 66.67 ft.

Given information:

The given function $h=-16t^2+72$ .

The given time range is 0 to 1 second.

Formula Used:

Integration power formula for the range $a\text{to}b$ is

  ${\displaystyle \underset{a}{\overset{b}{\int }}{x}^{2}dx=}{\frac{x^3}{3}|}_a^b=\frac{b^3}{3}-\frac{a^3}{3}$

Explanation:

Consider the given function $h=-16t^2+72$ .

Now, find height. Use the integration power formula for the range $t=0\text{to}1$ .

  $\begin{array}{c}h={\displaystyle \underset{0}{\overset{1}{\int }}\left(-16t^2+72\right)}dt\\ ={\left[-16\frac{t^3}{3}+72t\right]}_0^1\\ =-\frac{16}{3}+72\\ =66.67\end{array}$

Hence, the required height is 66.67 ft.

(c)

To check whether the distance vary of not at time $t=1\text{to}2$ .

The distance decreases.

Given information:

The given function $h=-16t^2+72$ .

The given time range is 1 to 2 second.

Formula Used:

Integration power formula for the range $a\text{to}b$ is

  ${\displaystyle \underset{a}{\overset{b}{\int }}{x}^{2}dx=}{\frac{x^3}{3}|}_a^b=\frac{b^3}{3}-\frac{a^3}{3}$

Explanation:

Consider the given function $h=-16t^2+72$ .

From 46b. at $t=0\text{to}1$ the height $h=66.67\text{ft}$ .

Now, find height. Use the integration power formula for the range $t=1\text{to}2$ .

  $\begin{array}{c}h={\displaystyle \underset{1}{\overset{2}{\int }}\left(-16t^2+72\right)}dt\\ ={\left[-16\frac{t^3}{3}+72t\right]}_1^2\\ =-\frac{128}{3}+144+\frac{16}{3}-72\\ =72-\frac{112}{3}\\ =34.67\end{array}$

Thus, at $t=1\text{to}2$ distance decreases.

Hence, the distance decreases.