Acceleration of Gravity

Acceleration of Gravity

Purpose: 1. To determine the acceleration of gravity for a freely falling object
2.  To gain experience using the computer as a data collector.

Equipment: Windows based computer, Lab Pro interface, Logger Pro software, motion detector, rubber ball, wire basket

Introduction: In this laboratory you will use the computer to collect some position (x) vs. time (t) data for a rubber ball tossed into the air. Since the velocity of an object is equal to the slope of the x vs. t curve each point in time. We will use both the x vs. t graph and the v vs. t graph to find the free fall acceleration of the ball.


Procedure: We logged in once again to use Logger Pro. We had to follow 8 steps that were given to us at the beginning of Lab.

• So we were given a blank graph position vs. time. The vertical scale (position axis) should be from 0 to 4 m while the horizontal scale (time axis) should be from 0 to 4. 
• We had to put the motion detector on the floor facing upward and place the wire basket (inverted) over the detector for protection from the falling ball. 
• We had to check properly to see if everything was working so we did a test run and pressed Collect to see if it was working. 
• Ideally your toss should result in the ball going straight up and down directly above the detector.

*Note: We would be repeating steps 4-6 for at least 5 trials. 

Trial #1:

Position vs. Time Graph (m/s)

This was our first position vs. time graph.

We got a parabolic graph which we only need the parabola part of the graph

We highlighted the parabolic part as seen in the graph. We then went to analyze/curve fit. 

We then used the equation x=At^2+bt+c

A=-4.651

B= 8.927

C=-2.892

We would only use a which we have to multiply it by 2 and we would get -9.302 m/s

that number would become part of my results table. 

Velocity vs. Time graph (m/s^2)

This graph represents velocity (m/s) vs. time (s).

The equation that we use is v=mt+b

Which I would I be using for my Results from Falling Body Experiment

This equation gave us the slope= -9.785 (m/s^2) which is shown above in our graph 

Trial #2:

Position vs, Time Graph (m/s)

I must say that this parabola is better than the first one.
It follows more the curve fit (wider).
The equation we used is x=At^2+Bt+C
We analyzed it once again like the first one.
A= -4.753
B=7.168
C=-1.124
We would multiply it by 2 which A= -9.506 that number would be
on Results table at the end.

Velocity vs. Time (m/s^2)

These numbers represent the velocity vs time graph.

The equation we used is v=mt+b

m which is the slope -9.610 (m/s^2) which is shown in our graph.

Trial #3:

Position vs. Time (m/s)

It was closer to the second graph that we got. 

It was because we had one person doing it, since she got the hang of it.

We followed the steps and used the equation

x=At^2+Bt+C

A=-5.078

B=7.817

C=-1.503

We multiplied it by 2 and got -10.156 for trial number 3.

Velocity vs. Time (m/s^2)

We basically highlight the part that is from the Top to the Bottom.

We use it to calculate v=mt+b

which m (slope) is equal -10.62.

Trial #4:

Position vs. Time (m/s)

As seen in previous graph they all seen pretty close to one another.

They all seem close to the curve fit. They are just couple inches apart from

each other.

x=At^2+Bt+C

A=-4.995

B=7.755

C=-1.384

We multiplied A once again by (2a)=-9.990.

Velocity vs. Time (m/s^2)

  

This graph represents Velocity vs. Time 

v=mt+b

m(slope)=-10.06

Trial #5:

Position vs. Time(m/s)

The Last and Final trial of this lab. It was one

that was closer to the first graph in Trial 1.

x=At^2+Bt+C

A= -4.767

B=6.967

C=0.8937

We than multiplied it by (2a)=-9.534

Velocity vs. Time(m/s^2)

The Last and Final Velocity 

The equation hasn't changed

v=mt+b

m(slope)=-10.10

All Five Velocity Graphs:

So after gathering all five trials we gathered all of our velocity vs. time graphs and put them together and this is what it looks like with all five trials. It looks good and it shows that our trials were all close.

Another note: 

Is that we used percent error= Measured-actual/actual*100 to find percent difference

As I mentioned through out my blog each number that I have on every graph corresponds to my table that is shown above. I know it looks kind of blurry because it was done with pencil. The only difference is that the percent difference is not on every graph. This was done after all the trials were recorded. 

Conclusion:

I learned that it takes time. I found that it was kind of hard for me get the percent difference. I actually talked to Dr. Haag about how the numbers should be put in which I actually took out the negative and entered it to the percent error= (9.80-#)/9.80*100. At first, I would get 2.08 number and then multiplied by 100 which would be 208.0 which I was wrong. I was able to fix it and got the right numbers. This lab was based on Position vs. Time (m/s) and then Velocity vs. Time (m/s^2). We would use a motion detector faced up protected by a wire basket. Once, we got it prepared we started trials. We would toss the ball up and then let it hit the wire basket which would give us the parabola and we would use that part for the curve fit. We used x=At^2+Bt+C for the graph Position vs. Time. For the second graph we used the equation v=mt +b and that would give us the slope and velocity. As you can see in my final results the very first graph the percent difference was the closest in my results for the velocity graph.