Acceleration of Gravity Edwin Gonzalez

Introduction

The purpose of this lab was to determine the acceleration of gravity for a freely falling object and to gain experience using the computer as a data collector. We used the computer to collect

some position (x) vs time (t) data for a ball tossed into the air. Since the

velocity of an object is equal to the slope of the position vs time curve, the computer can also construct the graph of velocity vs time by calculating the slope of position vs time at each point in time. We will use both the position vs. time graph and the velocity vs. time graph to find the free fall acceleration of the ball.

Data



Trial 1

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Trial 2

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Trial 3

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Trial 4

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Trial 5

 

Calculations

Position (m) vs Time (s)

Trial 1

x=At^2+Bt+C

x=(-4.863^2)+(7.7)t-(1.3)

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(2A-0.5*9.8)/ 0.5*9.8] x 100%

percent error = [((2*4.863)-9.8)/ 9.8)] x 100%

percent error = -2.04%

Trial 2

x=At^2+Bt+C

x=(-4.718^2)+(8.7)t-(2.2)

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(2A-9.8)/9.8] x 100%

percent error = [((2*4.718)-9.8)/ 9.8)] x 100%

percent error = -3.71%

Trial 3

x=At^2+Bt+C

x=(-4.703^2)+(7.7)t-(1.6)

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(2A-9.8)/9.8] x 100%

percent error = [((2*4.703)-9.8)/ 9.8)] x 100%

percent error = -4.02%

Trial 4

x=At^2+Bt+C

x=(-4.804^2)+(7.9)t-(1.8)

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(2A-9.8)/9.8] x 100%

percent error = [((2*4.804)-9.8)/ 9.8)] x 100%

percent error = -1.95%

Trial 5

x=At^2+Bt+C

x=(-4.708^2)+(7.821)t-(1.496)

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(2A-9.8)/9.8] x 100%

percent error = [((2*4.708)-9.8)/ 9.8)] x 100%

percent error = -3.91%

Velocity (m/s^2) vs Time (s)

Trial 1

y=mt+b

t = time

m = slope

b = y-intercept

y=9.885t+7.87

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(m-9.8)/9.8] x 100%

percent error = [(9.885-9.8)/ 9.8)] x 100%

percent error = 0.86%

Trial 2

y=mt+b

t = time

m = slope

b = y-intercept

y=9.892t+8.835

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(m-9.8)/9.8] x 100%

percent error = [(9.892-9.8)/ 9.8)] x 100%

percent error = 0.94%

Trial 3

y=mt+b

t = time

m = slope

b = y-intercept

y=9.306t+7.609

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(m-9.8)/9.8] x 100%

percent error = [(9.306-9.8)/ 9.8)] x 100%

percent error = -5.04%

Trial 4

y=mt+b

t = time

m = slope

b = y-intercept

y=9.886t+8.219

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(m-9.8)/9.8] x 100%

percent error = [(9.886-9.8)/ 9.8)] x 100%

percent error = 0.877%

Trial 5

y=mt+b

t = time

m = slope

b = y-intercept

y=9.449t+7.825

percent error = [(measured - actual)/(actual)] x 100%

percent error = [(m-9.8)/9.8] x 100%

percent error = [(9.449-9.8)/ 9.8)] x 100%

percent error = -3.58%

Results from Falling Body Experiment Trial gexp (2a)(m/s^2) % diff gexp (m)(m/s^2) % diff2 1 9.726 -2.04 -9.885 0.86 2 9.436 3.71 -9.892 0.93 3 9.406 -4.02 -9.306 -5.04 4 9.608 -1.95 -9.886 0.87 5 9.416 -3.91 -9.449 -3.58

Results from Falling Body Experiment
Trial	gexp (2a)(m/s^2)	% diff	gexp (m)(m/s^2)	% diff2
1	9.726	-2.04	-9.885	0.86
2	9.436	3.71	-9.892	0.93
3	9.406	-4.02	-9.306	-5.04
4	9.608	-1.95	-9.886	0.87
5	9.416	-3.91	-9.449	-3.58

Conclusions

In this lab we used logger pro software and logger pro interface to record the accelaration of gravity on a rubber ball. We simulated the ball falling on top of the motion detector with a wire basket on top of the motion detector (for protection) From this lab we concluded that the acceleration of gravity is 9.8(m/s^2) by taking  of the acceleration from the position vs time parabola graphs multiplied by 2, from the equation  by taking the average slopes of the velocity vs time graphs of y=mx+b where m = slope. The slope is the acceleration of gravity.