Chapter 2.1, Problem 44E

To determine

Describe the transformation of the graph of $f$ to obtain $g$ . Further, find the height from which the object should be dropped on the moon, so it hits the ground the same time as on Earth.

Answer to Problem 44E

The function $g$ is obtained when the function $f\text{ is vertically shrinked by a factor of }\frac{1}{6}$.

The height from which the object should be dropped on the moon, so it hits the ground the same time as on Earth is 1.67 feet.

Explanation of Solution

Given Information: The function $f\left(t\right)=-16t^2+10$ models the height of an object t seconds after it is dropped from a height of 10 feet on earth. The function $g\left(t\right)=-\frac{8}{3}{t}^2+10$ models the height of an object t seconds after it is dropped from a height of 10 feet on the moon.

Calculation:

The height of both the objects that fall from earth or moon is same. The only difference between both the functions is that of the prefix of $t^2$ . The function $g$ can be obtained by a vertical shift in the function of $f$ . That is,

  $\begin{array}{c}f\left(t\right)=-16t^2+10\\ =\left(\frac{1}{6}\right)-16t^2+10\\ =-\frac{8}{3}{t}^2+10\\ =g\left(t\right)\end{array}$

Hence, the function $g$ is obtained when the function $f$ is vertically shrinked by a factor of $\frac{1}{6}$ .

When the object reaches ground, height becomes 0. That is, $f\left(t\right)=0$ .

The time taken by the object to reach the ground on earth is,

  $\begin{array}{c}f\left(t\right)=0\\ -16t^2+10=0\\ -16t^2=-10\\ t^2=\frac{10}{16}\\ =\frac{5}{8}\mathrm{......}\left(1\right)\end{array}$

When the object on moon reaches ground, the height becomes 0. That is, $g\left(t\right)=0$ .

If the time taken by the object on moon to reach the ground at the same time as that of the object thrown on earth the time should be equal to that obtained in equation (1).

That is, $t^2=\frac{5}{8}$

Now substitute all these values in the model $g\left(t\right)=-\frac{8}{3}{t}^2+h$ to obtain theheight from which the object should be dropped.

Here, h represents the height from which the object should be dropped on the moon, so it hits the ground the same time as on Earth.

  $\begin{array}{c}g\left(t\right)=-\frac{8}{3}{t}^2+h\\ 0=\left(-\frac{8}{3}\right)\left(\frac{5}{8}\right)+h\\ h=\frac{5}{3}\\ \approx 1.67\end{array}$

Hence, the height from which the object should be dropped on the moon, so it hits the ground the same time as on Earth is 1.67 feet.