Free Fall Lab

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 1 2 3 4 5 6 7 8 9 10 Average
Small 0.585 sec 0.571 sec 0.567 sec 0.571 sec 0.571 sec 0.572 sec 0.571 sec 0.574 sec 0.576 sec 0.571 sec 0.573 sec Big 0.573 sec 0.568 sec 0.569 sec 0.569 sec 0.570 sec 0.569 sec 0.571 sec 0.563 sec 0.571 sec 0.570 sec 0.569 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