Free+Fall+Investigation

=Investigation : Free Fall=

Introduction
An object falls to the ground due to the forces of gravity. Gravity itself is constant, but there is a function that models the height of an object falling since speed acccelerates from zero to a constant speed. In this investigation, we will observe the time it takes for a block to fall to the ground from different heights. Using the materials learned in this unit, we will accurately develop a model describing the flight details of the object.

Purpose
To create a function that fits the relationship of the height of an object falling verses a selected time interval.

Equipment

 * A motion sensor
 * TI-83 Calculator
 * A small pillow or other soft object
 * A tape measure

Procedure

 * 1) Plug a motion sensor into a calculator that has been set up to collect data for a falling object. Let x represent time in seconds, and y represent height in meters.
 * 2) Set the block up to the desired height (1.8m, 1.5m) and start recording on the calculator. Straight after the recording starts, allow the block to fall right on top of the motion sensor. Look the collected data up in your calculator plot tables.
 * 3) Use the finite differences method to find the degree of polynomial function that models your data. Stop when the differences are nearly constant.

Our Recording Table: for height (m) of falling object in the alloted time (in secs)

 * Time (in seconds) || Height: 1.8 meters || Height: 1.5 meters ||
 * 0 || 1.8 || 1.402 ||
 * 0.054 || 1.728 || 1.365 ||
 * 0.108 || 1.631 || 1.304 ||
 * 0.161 || 1.512 || 1.218 ||
 * 0.215 || 1.371 || 1.11 ||
 * 0.269 || 1.21 || 0.99 ||
 * 0.323 || 1.031 || 0.88 ||
 * 0.376 || 0.761 || 0.375 ||
 * 0.43 || 0.107 || @ ||

@The object had already reached the ground at this time, so there is no need to record the height, knowing it is zero
 * The second numbers are in such a complex arithmetic sequence because the calculator took 94 even samples in 5secs.

Graphs
Object dropped from 1.8m

Object dropped from 1.5m using the same window

Finite differences:
Resulting Graph: It is quite obvious that the line that is found in finite differences does not match the data very well at all, therefore it is unlikely that we would use this as our model for the function.

We then tried using the second degree, and averaging out the differences. We however ignored the last difference as it was such an obvious outlier.

The average of the differences was -.033(6)

C=1.8 (the starting height) 2A=-.033(6) A=-.0168(3) A+B=-.072 B=-.0551(6)

The equation would be

Y=.0168(3)X(1/.054)^2+.0551(6)X(1/.054)+1.8

The graph looks like



Although it's not perfect, it's much closer to what we're looking for. Resulting Graph: Here too, the line does not meet up with the points, so we cannot use this formula either.

Now, since the finite differences technique fails to give us a solid formula to model our data, since the data points are either too evenly spaced out together, or too irregular, we use another method to find the model, i.e. the regression method on our calculator.

Regression
Quadratic Regression: For Height 1.8 meters y= -9.07x^2 + .4353x + 1.74 R^2= 0.974 Graph with line:

Cubic Regression: For Height 1.5 meters y= -36.79x^3 + 12.84x^2 + -2.3x + 1.425 R^2= 0.978 Graph with line:

Conclusion
Our final formulas for finding the formulas that models the height verses timespan are: 1.8m: y= -9.07x^2 + .4353x + 1.74 1.5m: y= -36.79x^3 + 12.84x^2 + -2.3x + 1.425 Y being height in meters, and X being time in seconds.