Free Fall
2/12/2013
Lab 1430
Free Fall
The difference of the outline procedure and the actual procedure is the use of the brass screw was not working in our set up. So we had to improvise and use our hand as the release mechanism as what we had seen this didn’t make difference from others results.
Drop Distance 50(cm)
Drop
Time(sec)
1
.306179
2
.310800
3
.304614
4
.311203
5
.298986
Drop Distance 100(cm)
Drop
Time(sec)
1
.419258
2
.417368
3
.420589
4
.416400
5
.430646
Drop Distance 150(cm)
Drop
Time(sec)
1
.516188
2
.504206
3
.495936
4
.515523
5
.502310
Drop Distance 200(cm)
Drop
Time (sec)
1
.623696
2
.616600
3
.618880
4
.628058
5
.602976
H(m)
.093756
.5
.177116
1
.256879
1.5
.38183
2
Stander Deviation
H(m)
Deviation
.5
.3063564
1
.4208522
1.5
.5068326
2
.618042
Equation Used
Percentage Error = (Abs(measured value-calculated value)/calculate value)*100%
What we can see from the results and the theory of the idea of the ball dropping is that the time it takes form 1 meter and 2 meters aren’t twice as large. What we can see is that it is an exponential increase in a small amount. In theory this is proven that the time is not double just because the distance is double. And that the acceleration without air resistance will always be constant -9.81 m/s squared
In question number 2 by ignoring air resistance would this tend to cause the measures value of g in this experiment to be larger or smaller. This question may be miss leading because in earth the value of g is a constant and does not change for the reason that it’s a constant it can’t be change to what in our will.
In question number 3 it’s asking in what point in time the ball value of acceleration will be the smallest. I believe that at the highest point of the ball the magnitude the balls acceleration will have the smallest. I believe that this is right because it will have no acceleration so the slop will be zero and will not be moving so no acceleration.
In question number 4 if the ball is thrown upwards and
Lab 1430
Free Fall
The difference of the outline procedure and the actual procedure is the use of the brass screw was not working in our set up. So we had to improvise and use our hand as the release mechanism as what we had seen this didn’t make difference from others results.
Drop Distance 50(cm)
Drop
Time(sec)
1
.306179
2
.310800
3
.304614
4
.311203
5
.298986
Drop Distance 100(cm)
Drop
Time(sec)
1
.419258
2
.417368
3
.420589
4
.416400
5
.430646
Drop Distance 150(cm)
Drop
Time(sec)
1
.516188
2
.504206
3
.495936
4
.515523
5
.502310
Drop Distance 200(cm)
Drop
Time (sec)
1
.623696
2
.616600
3
.618880
4
.628058
5
.602976
H(m)
.093756
.5
.177116
1
.256879
1.5
.38183
2
Stander Deviation
H(m)
Deviation
.5
.3063564
1
.4208522
1.5
.5068326
2
.618042
Equation Used
Percentage Error = (Abs(measured value-calculated value)/calculate value)*100%
What we can see from the results and the theory of the idea of the ball dropping is that the time it takes form 1 meter and 2 meters aren’t twice as large. What we can see is that it is an exponential increase in a small amount. In theory this is proven that the time is not double just because the distance is double. And that the acceleration without air resistance will always be constant -9.81 m/s squared
In question number 2 by ignoring air resistance would this tend to cause the measures value of g in this experiment to be larger or smaller. This question may be miss leading because in earth the value of g is a constant and does not change for the reason that it’s a constant it can’t be change to what in our will.
In question number 3 it’s asking in what point in time the ball value of acceleration will be the smallest. I believe that at the highest point of the ball the magnitude the balls acceleration will have the smallest. I believe that this is right because it will have no acceleration so the slop will be zero and will not be moving so no acceleration.
In question number 4 if the ball is thrown upwards and