Free Fall Lab

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Julie Kim
Free Fall Lab

Purpose:
to use collected data and the kinematics equations to determine the value of local gravity

Data:
height  161 cm(1 m/100 cm) = 1.61 m
mass of small ball  16.5 g
mass of big ball  28.0 g
12345678910Average
Small0.585
sec0.571
sec0.567
sec0.571
sec0.571
sec0.572
sec0.571
sec0.574
sec0.576
sec0.571
sec0.573
sec
Big0.573
sec0.568
sec0.569
sec0.569
sec0.570
sec0.569
sec0.571
sec0.563
sec0.571
sec0.570
sec0.569
sec

Analysis of Data:
1. Determine the average time of each set of ten drops.
See data table.
2. Use the average time, height and kinematics equations to determine local gravity.
Small Ball: ∆y = vit+(1/2)at2
1.61 = (0*0.573)+(1/2)a(0.5732)
a = 9.81 m/s2
Big Ball: ∆y = vit+(1/2)at2
1.61 = (0*0.569)+(1/2)a(0.5692)
a = 9.95 m/s2
3. Calculate the percent error between each calculated local gravity and the accepted value of gravity, 9.8 m/s2.
Small Ball: [(9.81–9.8)/9.8]*100 = 0.10%
Big Ball: [(9.95-9.8)/9.8]*100 = 1.53%
4. How do the drop times compare between the two different sized balls? Is this consistent with the concepts learned in class? Explain.
The drop times between the two different sized balls are very similar. The difference
between the two averages is only 0.004 seconds. This is consistent with the concepts
learned in class; we learned that all objects fall at the same rate. The two balls fell at very
close rates.

Error Analysis:
Air resistance could have caused the smaller ball to fall at a slightly slower rate. Also, the machine might not have been reset to zero before the ball was released. See #3 of the “Analysis of Data” section for percent error calculations.

Conclusion:
All objects, no matter what the volume or mass, fall at the same rate, as seen by the very close gravity values of the 16.5 g ball and 28.0 g ball. This lab was easy and quick to do. The only problem was...
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