Can someone please help me with this, I would appreciate it greatly thank you. I am not too good with spreadsheets and these instructions are so confusing.

Data Analysis: Use a spreadsheet to perform the following data analysis: 1. Compute the data reduction in Table 2 including the average of each data values, the standard deviation. Calculate the precision for each v value bv taking its corresponding standard deviation and dividing it by its average value. Also compute the value ofthe time squared lt"2} and place this in the Data Table. This will be used for graphical analvsis. 2. Make a plot of the "Average v" versus ''Time'\". This graph should be a scatter plot. Y is the dependent variable {yr-axis] and Time is the independent variable {x- axis}. Make sure vou include title, axis labels and units on your graph. This graph is showing how the position changes in time. The change of position in time is the velocitv. 3. Plot a graph of "Average v\" versus \"t\"2". Since we know that the equation of motion for an object in free fall starting from rest is v = 'A gt2, where g is the acceleration clue to gravity, plotting Average v vs t"2 should give us a straight line with slope gf2. [Average v vs tn2 is the equation of a parabola, which has the general form v = ax2}. 4. Determine the slope of this bv using the trendline function. The trendline is an Excel function that can extract a linear relationship from a set ofdata using the least-squares fit method. Please select "add trendline equation" so it shows on your plot from step #3. Remember, to use the proper significant figures for this slope. Since our data (3.! values} have 2 significant figures {while time onlv has 1, we are assuming that is accurate to greater the 2 significant figures}, we are allowed to use the minimum ofthe signicant figures for the data used in our reduction algorithm. (However, for some statistical functional analvsis, we can add one more significant figure.} 5. From the slope determined from the trendline, calculate the percent error of the value of g, the gravitational constant. (Use the accepted value of 'g' as 9.8 mfszl. [Percent Error = [{Experimental Determined Value Theoretical Value}! Theoretical 'v'alue ] * 100] (This is the same equation listed in the instructions} 6. Calculate the precision of your data bv taking the standard deviation at each time step and divide it by the average for that time step. MultiplyI the result by 100 to get a percentage. (In doing these calculations, remember to be mindful of the number of significant figures. Since our time is assumed to be precise, we onlv focus on the fall distance data. Anv calculation done with this data can onlvlr have the number ofdigits as the smallest number of significant figures in the calculation. The exception is statistical functions (average, standard deviation], where you can add one more digit. For example, ifwe are averaging the following numbers 1.4, 1.1, 1.4, 2.0, 3.0 together, we notice that each as 2 significant figures. Thus, for the average and standard deviation, we can have 3 significant figures. Ifhowever, we were averaging 14.3, 12.3, 2, 2.9 and 1.23, the smallest number of significant gures is 2 with one significant figure. That means, we can have 2 significant figures in our answer.) Data: A table of free-fall data will be provided for you similar to the table below: Data Table 1 - Dist. Dist. Dist. Dist. Dist. 7 rllml rzlml lm} Mm) 'rSlml Measurements were made of the distance offall (y) at each of the four preciseli,f measured times. This data this distance an object an object fell after these precise times. We are assuming at that each of the times, the object has not yet hit the ground and is falling under the influence of the gravitational force. Time 0.5 0.75 1.25 1.5 Distance Trial 1 (m) 1.32 2.97 5.25 8.25 11.7 Distance Trial 2 (m) 1.19 2.52 4.8 7.43 11.2 Distance Trial 3 (m) 1.12 2.6 4.72 7.45 10.34 Distance Trial 4 (m) 1.24 3.18 5.28 8.44 11.74 Distance Trial 5 (m) 1.12 2.81 4.17 8.72 11.46 Average Distance ("1) Standard Deviation ("1) Precision Stdev/Ave 0/5 Time Squared (5'5)