Chapter 4.5, Problem 36PPSE

Solution

a.

To Find: The constant $1700$ tell about the height of the object.

The object started at a height of $1,700$ feet.

Given: $h=-16t^2+1700$

Explanation:

Constant $1700$

tell about the height of the object in $h=-16t^2+1700$ :

The object started at a height of $1,700$ feet.

b.

To Find: The coefficient of $t^2$ tell about the direction the object is moving.

The coefficient of $t^2$ is negative, so it's moving down.

Given: $h=-16t^2+1700$

Explanation:

Coefficient of $t^2$:

As time increases, the height decreases since the coefficient is negative. As a result, the object is moving downward.

c.

To Find: The time at which the object is at a height $1000\text{ft}$

The time at which the object is at a height $1000\text{ft}$ is $6.6144\text{sec}$ .

Given: $h=-16t^2+1700$

Explanation:

The object height is determined by the equation:

  $h=-16t^2+1700$

Here, $t$ is the time in seconds.

Let, height of the object is $1000\text{ft}$

Find the time for which the height of the object $1000\text{ft}$

Substitute $h=1000$ in $h=-16t^2+1700$ :

  $\begin{array}{c}1000=-16t^2+1700\\ -16t^2=-700\\ t^2=\frac{700}{16}\end{array}$

Find $t$ :

  $\begin{array}{c}=\frac{175}{4}\\ =\sqrt{\frac{175}{4}}\\ =\frac{5\sqrt{7}}{2}\\ \approx 6.6144\end{array}$

Thus, the time at which the object is at a height $1000\text{ft}$ is $6.6144\text{sec}$ .

d.

To Find: The time at which the object is at a height $940\text{ft}$

The time at which the object is at a height $940\text{ft}$ is

Given: $h=-16t^2+1700$

Explanation:

The object height is determined by the equation:

  $h=-16t^2+1700$

Here, $t$ is the time in seconds.

Let, height of the object is $940\text{ft}$

Find the time for which the height of the object $940\text{ft}$

Substitute $h=1000$ in $h=-16t^2+1700$ :

  $\begin{array}{c}940=-16t^2+1700\\ -16t^2=-760\\ t^2=\frac{760}{16}\end{array}$

Find $t$ :

  $\begin{array}{c}t=\frac{95}{2}\\ =\sqrt{\frac{95}{2}}\\ \approx 6.892\end{array}$

Thus, the time at which the object is at a height $940\text{ft}$ is $6.892\text{sec}$ .

e.

To Find: The reasonable domain and range for the function $h$ :

The reasonable domain of the function is $0\le t\le 10.31$ .

Given: $h=-16t^2+1700$

Explanation:

Find the reasonable domain and range for the function $h$ :

When the object reaches the ground, it takes between zero and one second $\left(h=0\right)$

So, equate $h=0$ , then

  $\begin{array}{c}-16t^2+1700=0\\ 16t^2=1700\\ t^2=\frac{1700}{16}\\ t=10.31\end{array}$

Because the height function $h=-16t^2+1700$ has the domain of all real numbers, it excludes negative values within its domain.

So, the reasonable domain of the function is $0\le t\le 10.31$ .