Height (m) 1.58m 1,51m 1,413 1.34 m Height (m) Large Plastic Ball Data Time of Fall (s) Trial 2 Trial 1 Trial 3 0.590550158698 0.58728 6.58235 0.33905 0.57 685 015745 0.5751 s 0.5755 S 0.33 1252 0.5544505510d 0.55285 0.5527 S 0305554 0153395 0.5372 0176555 0.6ood & 0.36025 0584051665 0.51855 0.51815 10. 26845 Golf Ball Data Average Time (s) Time of Fall (s) Average Time Squared (S) Average Time (s) Average Time Squared (S) Trial 1 Trial 2 Trial 3 1.62m 0.57898 1.52M 0157898 0153905 015780 S 0.5787 S 0.33 49 5 0.56145 0.56155 0.5 6005 0.5 600$ 0.314752 1.41m 0.5443354166 $154155 054345 0.2932 5 1.30m 015390 s 0153905015289 0150305 0.5284 50 27925 S S 0.49245 0,4940 $ 0.493.45 0.4933 5 0 2 433 53 1.17 m (

13. Using Graphical Analysis, for the first ball make a graph of the Height (y-axis) vs. Time Squared (x-axis). According to the equations of constant acceleration, this graph should be a straight line. The slope of the line is 1/2 the acceleration of the ball. The program can calculate the slope of the line. Click the mouse on the graph. On the menu bar, click on Analyze. Select Linear Fit. A new window pops up. The program fits a line to your data. Read the value of the slope. Do this for each type of ball. Calculate the free fall acceleration of the ball by multiplying the slope by 2 . Remember the slope has units. 14. Compare to the accepted value of free fall acceleration: g=9.80m/s2 by calculating the percent error. How well does your value of the free fall acceleration agree with the accepted value of free-fall acceleration, g=9.8m/s2? 15. Were the accelerations of each ball the nearly the same? Which one was the closest to 9.80m/s2 ? Can you think of a reason why the accelerations were not the same