Table 1: height transit transit y [m] ti transit average transit average me t. (
ID: 3210038 • Letter: T
Question
Table 1: height transit transit y [m] ti transit average transit average me t. (s) time t, (s) time t, (s) transit time time squa 6 2 3 .38". 3g41 5 6 Results: (a) On the graph paper provided make a graph of height y (on y axese on both Choose an appropriate scale on both average transit time squared (on x axes). axes to use at least 75% of the graph paper. b) Draw the best fit line (line in between data point without hitting any data point) and find the slope of the best fit line. To calculate slope you have to consider two additional points that lie on the best fit line, not the data points. Show your points on the graph and your calculation to get credit. (c) From the slope, determine gravitational acceleration g. How is slope related to gravity g. HINT: Write the kinematic equation that you plot and compare it to the equation of a line y=m x where m is slope. Slope= c) slope of the best fit line (write unit) d) measured g (from slope) = a) Calculate the percentage error with respect to the standard v % percentage error = Ig(from slope)-9.81/9.8 x100- ( write unit) alue of 9.8m/sExplanation / Answer
a) Graph of y vs average of X is given below:
The points that lie on regression line approximately are (0,0) and (2,9)
b) Slope of regression line =
= 4.521
The regression line is y = 0.035+4.521x
The y intercept is almost negligible.
Hence regression line is y = 4.521 x
c) Compare with y =mx
For falling body s= ut+1/2 at^2
= 1/2 at^2
Hence 1/2 a = slope
Or g = 2(slope)
d) Measured g from slope = 4.521(2) = 9.024
e) Percentage error = |9.024-9.8|(100)/9.8 = 7.918
Hence percentage error = 7.918
y x=x bar ^2 y^2 x^2 1 0.2094 1 0.043848 0.6 0.1252 0.36 0.015675 0.7 0.1475 0.49 0.021756 0.8 0.1665 0.64 0.027722 0.9 0.1975 0.81 0.039006 Mean 0.8 0.16922Related Questions
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