Galileo studied motion by rolling balls down ramps. He rolled a ball down a ramp with a horizontal shelf at the end of it so that the ball was moving horizontally when it started to fall off the shelf. The top of the ramp was placed at different heights above the floor (that is, the length of the ramp varied), and Galileo measured the horizontal distance the ball traveled before it hit the floor. Here are Galileo’s data. (We won’t try to describe the obsolete seventeenth-century units Galileo used to measure distance and height.)

$\begin{array}{cc}\text{Distance}& \text{Height}\\ 1500& 1000\\ 1340& 828\\ 13328& 8000\\ 1172& 600\\ 800& 300\\ & \end{array}$

(a) Make an appropriate scatterplot for predicting horizontal distance traveled from ramp height. Describe what you see.

(b) Use transformations to linearize the relationship. Does the relationship between distance and height seem to follow an exponential model or a power model? Justify your answer.

(c) Perform least-squares regression on the transformed data. Give the equation of your regression line. Define any variables you use.

(d) Use your model from part (c) to predict the horizontal distance a ball would travel if the ramp height was $700$. Show your work.

$\begin{array}{|ll|}\hline \text{Distance}& \text{Height}\\ 1500& 1000\\ 1340& 828\\ 1328& 800\\ 1172& 600\\ 800& 300\\ \hline\end{array}$

The scatterplot is swinging higher, hence the direction is positive.

Because there is no straight line connecting the ends, the form is curved.

Because all spots in the same pattern are so close together, the strength is solid.

$\begin{array}{|ll|}\hline \text{Distance}& \text{Height}\\ 1500& 1000\\ 1340& 828\\ 1328& 800\\ 1172& 600\\ 800& 300\\ \hline\end{array}$

Exponential Model

Power Model

The height logarithm is in the horizontal dimension in the Exponential Model, the distance logarithm is in the y axis in the Power Model, and the height logarithm is in the horizontal axis in the Power Model.

The Power Model is used to model the relationship between the variables, with the most linear version shown in the accompanying scatterplot.

$\begin{array}{|ll|}\hline \text{Distance}& \text{Height}\\ 1500& 1000\\ 1340& 828\\ 1328& 800\\ 1172& 600\\ 800& 300\\ \hline\end{array}$

The height and distance values are logarithmic.

$\begin{array}{|llll|}\hline \text{Height}\left(X\right)& \text{Distance}\left(Y\right)& XY& XX\\ 6.907755& 7.31322& 50.517932& 47.71708\\ 6.719013& 7.200425& 48.379749& 45.14514\\ 6.684612& 7.191429& 48.071913& 44.68404\\ 6.39693& 7.066467& 45.203695& 40.92071\\ 5.703782& 6.684612& 38.12757& 32.53313\\ \sum X=32.412092& \sum Y=35.456153& \sum XY=230.30086& \sum X^2=211.0001\\ \hline\end{array}$

$a=\frac{\sum Y·\sum X^2-\sum X·\sum XY}{n\sum X^2-{\left(\sum X\right)}^2}=3.751$

$b=\frac{n\sum XY-\sum X.\sum Y}{n\sum X^2-{\left(\sum X\right)}^2}=0.515$

$y=a+bx$

role="math" localid="1652861160820" $y=3.751+0.515x$

The equation from part(c)

$y=3.751+0.515x$

Using the value of $x=700$in the component equation (c)

$$\begin{gathered} \text{ln}\hat{y}=3.751+0.515\ln \left(700\right) \\ =7.12481 \end{gathered}$$

Now we take the Exponential of the above equation

$$\begin{gathered} \hat{\mathit{y}}=e^{7.12481} \\ =1242.41 \end{gathered}$$