Chapter 8.2, Problem 39E

a.

To determine

To write: the equations that represents the height h (in feet) as a function of time t (in seconds) for both the acorns.

Answer to Problem 39E

Acorn 1: $h=-16t^2+45$

Acorn 2: $h=-16t^2+32$

Explanation of Solution

Given information:

The first acorn falls 45 feet.

The second acorn falls 32 feet.

Concept used:

The height (in feet) of a falling object t seconds after it is dropped from an initial height of $s_0$ (in feet) is given by the function:

  $h\left(t\right)=-16t^2+s_0$   ..... (1)

Calculation:

The height h (in feet) for the first acorn as a function of time t (in seconds) is found by substituting $s_0=45$ in equation (1),

  $h=-16t^2+45$

The height h (in feet) for the second acorn as a function of time t (in seconds) is found by substituting $s_0=32$ in equation (1),

  $h=-16t^2+32$

b.

To determine

To describe: how graphs of the two equations are related.

Answer to Problem 39E

The graph which represents the height of the first acorn is a vertical translation 13 units up of the graph representing the height of the second acorn.

Explanation of Solution

Given information:

The equations that represents the height h (in feet) as a function of time t (in seconds) for both the acorns are:

Acorn 1: $h_1=-16t^2+45$

Acorn 2: $h_2=-16t^2+32$

Concept used:

Vertical translation:

The graph of $g\left(x\right)=f\left(x\right)+k$ is a vertical translation of the graph of f .

The vertical shift is k units up if $k>0$ and k units down if $k<0$ .

Consider the equation representing the height of first acorn:

  $\begin{array}{c}{h}_1=-16t^2+45\\ =-16t^2+32+13\\ =h_2+13\end{array}$

It can be observed that $h_1\left(t\right)=h_2\left(t\right)+13$

So, the graph of $h_1$ is a vertical translation of the graph of $h_2$ .

Now, as $13>0$ the shift is 13 units up and so the graph which represents the height of the first acorn is a vertical translation 13 units up of the graph representing the height of the second acorn.