Question

Displacement (meters)	Time (seconds)
Trial 1 (0.70 m)
0.695 m	0.373 s
0.695 m	0.374 s
0.695 m	0.373 s
0.695 m	0.373 s
Average Displacement = 0.6951 m	Average Time = 0.3733 s
Trial 2 (0.60 m)
0.598 m	0.345 s
0.595 m	0.345 s
0.595 m	0.345 s
0.598 m	0.346 s
Average Displacement = 0.5960 m	Average Time = 0.3453 s
Trial 3 (0.50 m)
0.495 m	0.314 s
0.495 m	0.315 s
0.495 m	0.314 s
0.495 m	0.314 s
Average Displacement = 0.4950 m	Average Time = 0.3143 s
Trial 4 (0.40 m)
0.395 m	0.281 s
0.3948 m	0.279 s
0.3941 m	0.280 s
0.395 m	0.279 s
Average Displacement = 0.3947 m	Average Time = 0.2798 s
Trial 5 (0.20 m)
0.200 m	0.197 s
0.1975 m	0.197 s
0.198 m	0.197 s
0.199 m	0.196 s
Average Displacement = 0.1986 m	Average Time = 0.1968 s

Physics Measurement and Free Fall Lab. The equations I have are: y = 4.9318x^2 - 0.0061x + 0.0101 for 'Displacement vs. Time' graph and y = 4.9168x +0.0097 for 'Displacement vs Time^2' graph.

How do the experimental values of "g" compare with the accepted value (g = 9.80 m/s^2) for the acceleration of gravity? Give a mathematical comparison.

Answer 1

For the free fall, the distance of fall at any time is given by .

The above equation resembles the equation of a straight line , with zero y-intercept.

The graph is drawn with Average displacement on Y-axis and (Average time)² () on X-axis.Which is a straight line with slope .

Comparing the straight line equation with distance equation ,

Using the given displacement - time² graph  

Slope of the straight line is

So

Theoretical value of acceleration due to gravity

Percent error in acceleration due to gravity is