Ross+and+Chris

Lab: What is the Acceleration of a Falling Body? Ross and Chris date

Objective: Find the acceleration of a falling body

Hypothesis- The acceleration of the object will be 980 cm/s2

Procedure- Slide a piece of ticker tape through a ticker tape machine mounted on the side of a door. Attach an object at the end of the ticker tape that is through the machine. Set the machine to 60 Hz. Turn on machine and drop object simultaneously. After turning off machine and collecting ticker tape, number each dot starting with 0. Measure the displacement of each dot from dot 0. Record results on a chart and make a graph.

Data and Graph: great results! Calculations: Percent Difference: __abs(your value - class average)__ *100 (your value + class average)/2

__abs(916.5-874.43)__ *100= **4.70%** (916.5+874.43)/2

Percent Error: __abs(Theoretical Value- Experimental Value)__ Theoretical value

__abs(981-916.5)__ *100= **6.57%** 981

d=**A**t^2+__B__t d=__initial velocity * time__ + **1/2acceleration*time**^2 **458.25**x^2 - __14.18__x 458.25=1/2*a a= 916.5 m/s 2

Discussion Questions-

1. Does the shape of your graph agree with the expected graph? Why or why not?

Yes, it is originally curved, then becomes more sloped towards the end. It agrees with the expected graph since it is originally sloped as it is accelerating. However, once it becomes linear, it is obvious that the object has reached constant velocity. However, the initial velocity should be zero, thus making it only the squaring of the acceleration.

2. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) The class averaged 872.43 cm/s2 to our 916.5 cm/s2. This has a percent difference of 6.81%, so we were relatively accurate compared to the class, as the Percent difference would be significantly less if one of the outliers were not included.

3. Did the object accelerate uniformly? How do you know?

Yes, it maintained a curved line before eventually becoming a line once free fall was reached. but how do you know it was UNIFORMLY accelerating

4. What should the velocity-time graph of this object look like?

At first, it would start at zero, and keep on increasing with a slope of 980 cm/s2. Then, after it reaches the constant velocity it would even out as it would have a slope of 0. when does it reach constant?

5. Write down the expected equation of the line from this v-t graph (use specific information from your x-t graph). y=916.5x

6. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

The weight of the object, no! and a subtle, unintentional "push" on the object could be reasons for it being higher. However, if the weight did not have enough time to reach free fall it would be lower than it should be. Finally, friction was caused by the tape rubbing against the machine, so that would also make it lower. I think you are using the term "freefall" instead of "terminal velocity". Objects need to fall for a long time and far distance before they reach terminal velocity... not an issue here.

Conculsion:

The results from this freefall experiment supported our hypothesis. We expected the weight to fall with constant acceleration, thus increasing velocity. However, towards the end of the graph, we noticed that the trendline started to straighten out, no it is parabolic all the way through (look at your R^2!) thus the velocity began to become constant. This implies that there must be a speed at which an object falling due to gravity cannot increase its velocity past. That speed is 9.8 m/s. <- No! you are making this up... if you really think this, before you write it out, you should look it up to verify. In this case, you are way off. we did not get 980 cm/s as we expected, however there are numerous factors that can account for this. First, friction. the friction between the sparker and the spark tape may have created drag that would have caused the velocity not to reach its maximum potential. Second, human error. <- this is not a sentence! When measuring the distances between the dots on the spark tape, there is a possibility we may have made a mistake in judgement. A good example of some implications of this data would be to examine the military. In both parachute troopers and delivering foreign aid, people and crates are jettisoned from a plane and are able to land safely on the ground with the use of parachutes. Without parachutes, the velocity of the crates and people would reach 9.8 m/s. Thus, if the crates hit the ground at this speed, they would break, probably rendering their contents useless. The military uses parachutes to slow the velocity of the freefall, protecting the supplies. good application, but you keep using 9.8 as a speed not the acceleration. You need to make sure you know the difference between speed and acceleraiton. It is extremely important here.