Chapter 9.2, Problem 25PPS

a.

To determine

To write: The functions that model the height $h$ of each rock after $t$ seconds.

Answer to Problem 25PPS

The function $h=-16t^2+300$ is model the height of first rock after $t\mathrm{seconds}$ and the function $h=-16t^2+700$ is model the height of first rock after $t\mathrm{seconds}$ .

Explanation of Solution

Given information: A rock falls from a cliff $300\mathrm{feet}$ above the ground. At the same time another rock falls from a cliff $700\mathrm{feet}$ above the ground.

Calculation: A rock falls from a cliff $300\mathrm{feet}$ above the ground. At the same time another rock falls from a cliff $700\mathrm{feet}$ above the ground.

The height of the rocks can be modeled using the equation $h=-16t^2+h_0$ where $h_0$ is the initial height of the rock. For the first rock, it has an initial height of $300\mathrm{feet}$ so $h_0=300$ . Its equation is then, $h=-16t^2+300$ .. For the second rock, it has an initial height of $700\mathrm{feet}$ so $h_0=700$ . Its equation is then, $h=-16t^2+700$ .

Thus, the function $h=-16t^2+300$ is model the height of first rock after $t\mathrm{seconds}$ and the function $h=-16t^2+700$ is model the height of first rock after $t\mathrm{seconds}$ .

b.

To determine

To calculate: How much sooner will first rock reach the ground.

Answer to Problem 25PPS

The first rock reaches the ground about $2.3\mathrm{seconds}$ sooner than the second rock.

Explanation of Solution

Given information: A rock falls from a cliff $300\mathrm{feet}$ above the ground. At the same time the another rock falls from a cliff $700\mathrm{feet}$ above the ground.

Calculation: A rock falls from a cliff $300\mathrm{feet}$ above the ground. At the same time the another rock falls from a cliff $700\mathrm{feet}$ above the ground.

When the rock hit the ground their height will be $h=0$ .

Put $h=0$ in both equations :

  $\begin{array}{l}h=-16t^2+300\\ 0=-16t^2+300\\ 16t^2=300\\ t^2=\frac{300}{16}\\ t=\sqrt{\frac{300}{16}}=\frac{10}{4}\sqrt{3}=\frac{5}{2}\times 1.73=5\times 0.86=4.3\mathrm{seconds}\end{array}$

  $\begin{array}{l}h=-16t^2+700\\ 0=-16t^2+700\\ 16t^2=700\\ t^2=\frac{700}{16}\\ t=\sqrt{\frac{700}{16}}=\frac{10}{4}\sqrt{7}=\frac{5}{2}\times 2.64=5\times 1.32=6.6\mathrm{seconds}\end{array}$

Time taken the first rock reaches the ground about sooner than the second rock is $6.6-4.3=2.3$ .