Free Fall Lab determination of g

Purpose: the purpose of this lab is to show whether or not an object at free fall will accelerate to the ground at 9.8 m/s^2 due to gravity and in the absence of all other external forces.




This apparatus is used to mark the position of the free falling body during the same intervals. A spark generator marks the position on a spark sensitive tape giving a record of the fall.

To fulfill the purpose of the lab we are using the data we collected and putting it in an excel worksheet. In the excel worksheet there will be 5 columns of data imputed into the spreadsheet. The first column is the Time column at every 1/60 of a second. The second column is a Distance column in this column we chose a dot to be the t=0 and we measured the distance between the dot and the following dot and kept doing so until we felt like we had enough data points. The third column represents the change in position between the dots. The fourth column we calculated the mid interval time. Finally in the Fifth equation we calculated the mid interval speed.

 

 

 

 

 

 

Using the information from the excel columns D and E we created a scatter plot of velocity and time. The x- axis represents the time and the y-axis represents the velocity. This graph is linear showing that the acceleration of the object is constant and not increasing or decreasing.

 

 

 

 

For the following graph we used columns A and B from the excel data to create a scatter plot for position and time. Position being the y-axis and time being the x-axis. For this graph we had to add a trendline with a polynomial fit because a linear trendline would indicate that the velocity is constant but because the velocity is increasing the graph is not a straight line.

 

 

In the end we can use either graph to find out the acceleration due to gravity keep in mind there is going to be error so the exact value you will get might not be 9.8 m/s^2

 

 

Questions / Analysis:

 

1. To get the acceleration due to gravity from the velocity/time graph we just take the derivative of the equation of our graph y = 950.53x + 50.658 which becomes a = 950.53 which is in centimeters so converted to meters it is 9.5053 m/s^2 and the accepted value is 9.8 m/s^2 which is pretty close considering the equipment we used is not the best and there are other factors that could have caused error.
2. To get the acceleration due to gravity from the position/ time graph we just take the second derivative of the equation y = 472.88x^2 + 50.937x + .0373. The second derivative gives the acceleration to be a = 945.76. This ends up being 9.4576 m/s^2 which is also a good value compared to the accepted value of 9.8 m/s^2