E03 Free Fall Lab_Legaspi-1

PHYSICS 100A FREE FALL DATA SHEET INSTRUCTIONS = This page assumes you’ve already read and as a group understand the instructions full instructions on obtaining the data for this laboratory. Each lab partner is expected to create three short videos of them dropping the ball to analyze using Tracker. Be sure to include an object of known height in your videos for calibration. (Obtain one screenshot of each partner dropping the tennis ball == participatory screenshots) Analyze videos in consultation as a group. (Obtain screenshots of EACH Tracker showing the graph and the values for the fit.) Once all videos have been analyzed, select one person in your group to compile the data and enter it in the CURVE FIT DATA TABLE. As a group answer the questions in the space provided. Include pictures of the screenshots along with your reports.

PHYSICS 100AL EXPERIMENT 3 DATA AND REPORT CURVE FIT DATA - Partner Name Trial A g exp = |2A| Sofia 1a -510 cm/s 2 1020 cm/s 2 Emily 1b -550 cm/s 2 1100 cm/s 2 Sofia 2a -497 cm/s 2 994 cm/s 2 Emily 2b -464 cm/s 2 928 cm/s 2 Sofia 3a -472 cm/s 2 944 cm/s 2 Emily 3b -473 cm/s 2 946 cm/s 2 g average 988.6 cm/s 2 s g 64.7 cm/s 2 Analysis Questions. 1. Compare your value of g with the accepted value of g = 979.2 cm/s 2 . a. Is the accepted value within one standard deviation from your average value? Yes, the results of our average value are within one standard deviation from the accepted value. As shown in the graph, our standard deviation is 64.7 cm/s 2 , the average value is 988.6 and the accepted value is 972.2 cm/s 2 . To determine whether or not it is within one standard deviation we can calculate the upper bound [972.2 cm/s 2 + 64.7 cm/s 2 = 1036.9 cm/s 2 ] and the lower bound [972.2 cm/s 2 - 64.7 cm/s 2 = 907.5 cm/s 2 ]. Going back to our average value we can see that the value 988.6 cm/ s2 falls within the range of 907.5 cm/s 2 and 1036.9 cm/s 2 . b. Calculate the percent error in your measured value of g? The percent error for the measured value of g is 0.95% (972.2 – 988.6) / 988.6 x 100 = 0.95% 2. Consider the values of g obtained by each lab partner. Were any partner’s results significantly better or worse than the others? Why do you think this is so? Based on the experiment and calculations, partner A had numbers ranging from 944 cm/s 2 to 1020 cm/s 2 which results to having a standard deviation of 38.6 cm/s 2 .

Trial 1B Trial 2B Trial 3B

Sample Tracker screen shot: Trial 1A Trial 2A

Trial 2B Trial 3B