Determining the Value of g

1. Ana Leyva
3/1/17
2. With this experiment we were trying to calculate the acceleration of a free falling object.
3. By taking the marks that the spark generator created on a piece of spark sensitive tape we plotted the distance between the marks into an excel chart. using the values on the excel chart we then created a velocity vs time graph and a distance vs time graph which both helped us find the acceleration of the free falling object. The slope of our velocity vs time graph will give us the acceleration and the ax^2 value of the polynomial fit of the distance vs time graph will also give us a value for the acceleration when you multiply it by two.

 4. The procedures for this experiment were pretty simple we had to measure the distance between the marks that were created on the paper. The dots on the paper corresponded to the the position of the falling mass every 1/60th of a second. After that we had to enter the distance into an excel spreadsheet. Once that information was all enter we calculated the change of x, the mid interval time, and the mid interval speed on that same excel sheet. We did this differently for each column. For the time column we started at zero and increased each one after that by 1/60 of a second. For the distance column we started at zero and input all of our measurement after calculating them with a ruler. For the change in x column we subtracted the distance from the row we were on from the row before it. The mid time interval was calculated by adding the time from the row we were on by 1/120. Finally, the mid-interval speed was calculated by dividing the change in x by 1/60. After all the values were calculated we created a velocity vs time graph and a position vs time graph. By creating a linear fit and a polynomial fit of order 2 we were able to get the values of acceleration from the two graphs. For the velocity time graph the acceleration was given by 960.65 and for the distance vs time graph the acceleration was given by 484.485. However, you have to multiply the acceleration for the distance vs time graph by two because the acceleration is given by .5ax^2 so the acceleration for that graph is 968.97.

7. The reason why we graphed a position vs time graph and a velocity vs time graph was because those two graphs were going to help us find the acceleration of the free falling object.
8. The acceleration that we got from our velocity vs time graph was 960.65 cm/s^2 and for our position vs time graph we got 968.97 cm/s^2. Both of these values are less than the expected value which was 981.0 cm/s^2. The absolute difference was -20.35 cm/s^2 and -12.03 cm/s^2. The reason why our answer wasn't what we expected it to be could be due to air resistance which wasn't taken into account when doing this experiment.
Questions:
1.

2. In order to get the acceleration due to gravity  from my velocity vs time graph I had to get a linear fit from my data point and get the slope of that line. In this case the slope for that line was 960.65 cm/s^2.
3. The acceleration due to gravity can be found using a polynomial fit of the second degree for the position vs time graph. The formula for the polynomial fit is y = y_initial + v_initial*t +.5at^2 from this equation we set 484.485t^2 equal to .5at^2 and we get a= 968.97