Chapter 9.3, Problem 29PPS

(a)

To determine

Write two functions that models the height of the rocks after t seconds.

Answer to Problem 29PPS

The two functions are: $h=-16t^2+300\text{and}h=-16t^2+700$ .

Explanation of Solution

Given:

A rock drops from a cliff 20, 000 inches above ground. Another rock drops from a cliff 30,000 inches above the ground.

Concept Used:

The formula $h=-16t^2+h_0$ can be used to approximate the number of seconds t it takes for the rock to reach height h from an initial height of $h_0$ in feet.

Calculation:

For the first rock, the initial height is 300 ( $h_0=300$ ) feet.

Therefore, the function for the first rock is $h=-16t^2+300$

For the 2^nd rock, the initial height is 700 ( $h_0=700$ ) feet.

Therefore, the function for the second rock is $h=-16t^2+700$

Thus, the two functions are: $h=-16t^2+300\text{and}h=-16t^2+700$ .

(b)

To determine

Find rock will reach the ground first.

Answer to Problem 29PPS

The first rock will reach the ground 2.3 seconds sooner than the second rock.

Explanation of Solution

Given:

A rock drops from a cliff 20, 000 inches above ground. Another rock drops from a cliff 30,000 inches above the ground.

Concept Used:

In order to figure out how long it takes each rock to reach the ground, set the height equal to 0 for both the equations.

Calculation:

In order to figure out how long it takes each rock to reach the ground, set the height equal to 0 for both the equations.

  $\begin{array}{l}h=-16t^2+300\text{original}\text{equation}\text{.}\\ 0=-16t^2+300\text{replace}h=0\\ 16t^2=-16t^2+16t^2+300\text{add}16t^2\text{to}\text{each}\text{side}\text{.}\\ 16t^2=300\text{simplify}\\ \frac{16t^2}{16}=\frac{300}{16}\text{divide}\text{each}\text{side}\text{by}16\\ t^2=18.75\\ \sqrt{t^2}=\sqrt{18.75}\text{take}\text{the}\text{positive}\text{square}\text{root}\text{of}\text{both}\text{sides}.\\ t\approx 4.3\end{array}$

It takes the first rock about 4.3 seconds to reach the ground.

  $\begin{array}{l}h=-16t^2+700\text{original}\text{equation}\text{.}\\ 0=-16t^2+700\text{replace}h=0\\ 16t^2=-16t^2+16t^2+700\text{add}16t^2\text{to}\text{each}\text{side}\text{.}\\ 16t^2=700\text{simplify}\\ \frac{16t^2}{16}=\frac{700}{16}\text{divide}\text{each}\text{side}\text{by}16\\ t^2=43.75\\ \sqrt{t^2}=\sqrt{43.75}\text{take}\text{the}\text{positive}\text{square}\text{root}\text{of}\text{both}\text{sides}.\\ t\approx 6.6\end{array}$

It takes the second rock about 6.6 seconds to reach the ground.

Since

  $6.6–4.3=2.3$ , the first rock will reach the ground 2.3 seconds sooner than the second rock. Thus, the first rock will reach the ground 2.3 seconds sooner than the second rock.