Acceleration due to gravity lab

Observations/Results:

While completing this experiment I noticed that the surface of the basketball was relatively rough (implications are further discussed in the in the Discussion section). I noticed that the acceleration of the ball increased as it was dropped (refer to Figure 1).  I also noticed that in some trials the ball did not travel in a straight line, it might have moved at an angle. In fact for Height #1, the distance travelled in Trial#2 was further than the distance travelled in Trial#1 and Trial#3(shown in Table I). I also noticed the same thing in Height#2, where in Trial #3 the distance travelled by the ball was more than 20cm greater than the distance the ball traveled in other (shown in Table II).  I noticed that for both heights, the velocity did not increase at equal rates (shown on Table II).  I noticed that on the velocity time graph (Figure 1), the slope which displays the acceleration due to gravity for Height #1 was 7.9422m/s2 with an R2=0.8573 , and for Height #2 was 12.851m/s2 with an R² = 0.6591, rather than both heights having a slope of 9.81m/s2. In the position time graph (Figure 2), I used a polynomial (of power 2) to model the best fit curve which gave me the equation of the line in the format of y=ax2+bx which I used to relate to the equation d= v1t+ ½ gt2 and found g, assuming the initial velocity was not zero. From this I found that the position time graph was a more accurate depiction of the acceleration of the ball as the values obtained for g were much closer to 9.81m/2.

Results/ Calculations:

Obtained from the Velocity Time Graph
Height#1: 7.9422m/s2 with an R2=0.8573
Height#2: g=12.851m/s2 with an R² = 0.6591

Using Formula
Height #1
d= v1t+ ½ gt2
g=2d/t2
g=2(3.57031)/ 0.93602
g=8.1505 m/s2

Height #2
d= v1t+ ½ gt2
g=2d/t2
g=2(2.193322)/ 0.93602
g=5.0070 m/s2

Obtained From Position Time Graph
Example: If Y=ax2+bx then from the formula d= v1t+ ½ gt2, you can assume that ½ g=a
Therefore:
½ g=a
2 x ½ g= 2a
g= 2a

Acceleration Due to Gravity Found (From Position Time Graph-Figure #2) 
Height#1
Y=ax2+bx
y = 4.2664x2 + 0.3952x
d= v1t+ ½ gt2
½ g=a
½ g=4.2664m/s2
g= 2(4.2664)m/s2
g= 8.5328 m/s2, or g=9m/s2

Height#2
Y=ax2+bx
y = 4.0709x2 + 1.6316x
d= v1t+ ½ gt2
½ g=a
½ g=4.0709 m/s2
g= 2(4.2664)m/s2
g= 8.1418m/s2, or g=8m/s2

Discussion

The graph of velocity time shows a fairly large error through the lines of best fit (R2=0.6581, and R2=0.8573). The values on the velocity time graph could have included the air resistance which could create a frictional force along the surface of the ball, especially due to the fact that it is a relatively rough ball, explaining why the acceleration due to gravity is 9.81m/s2 exactly. When considering that the acceleration of falling objects is 9.81m/s 2, we ignore air resistance. The assumption that the ball is dropped at no initial velocity may have been inaccurate, as the person dropping it could have exerted extra force, and pushed down to let go, rather than just letting go of the ball which would result in a high “g” value, as shown for Height #2 on Figure 1. For the position time graph, a more accurately fit quadratic model is obtained, (R2=0.9944 for Height #2, and R2= 0.9988 for Height#1) to find “g”, which results in 8.5328 m/s2 for Height#1, and 8.1418m/s2 for Height#2. These values are quite close to 9.81m/s2, but are still lower due to air resistance and the frictional force of the ball counteracting acceleration. These values seem to be more accurate than those obtained from the velocity time graph(Figure 1) because they eliminate the error of assuring V1=0. As expected the effect of air resistance is greater on the lower height (Height#2). Experimental error or weakness in this experiment is that we did not account for air resistance playing a factor in our results, the fact that the rough surface of the ball could have an effect on results, and that the way the ball is dropped can affect the results greatly. To improve this experiment for next time, in order to obtain much more accurate results I would need a ball with a very smooth surface being dropped in a vacuum. Overall, using the results obtained from the Position-Time graph(g=8.5328 m/s2 for Height#1, and 8.1418m/s2 for Height#2), I can come to the conclusion that the height that an object is dropped does not affect the acceleration due to gravity of that object, when there is no air resistance. I can come to this conclusion due to the fact that the difference between the value for Height #1 and Height#2 is not large. I can also come to the conclusion that in order for the acceleration due to gravity of an object to be 9.81m/s2 , the velocity must increase at a constant rate of 9.81m/s for each time interval, and that factors such as air resistance and friction can affect acceleration due to gravity greatly.

Bibliography

Chesick, Elizebeth. “Acceleration Due to Gravity” http://www.haverford.edu/educ/knight-booklet/accelarator.htm

The Physics Classroom “Freefall and the Acceleration of Gravity”www.physicsclassroom.com/Class/1Dkin/u1l5b.cfm