Tomas BilkauskasExtended essay in Physics 002390-0006
Extended essay in physics
Investigating air resistance phenomenon simulated in water and determining how force of resistance is dependent from objects velocity.
Word count:3656
Abstract
This essay investigates the how friction force is dependent from velocity of a falling object in water and whether the air resistance theory applies to falling objects in other mediums such as air. The experiment conducted was dropping a ball in a glass vessel full of water and using camera capture tools to measure the displacement versus time data that was later used graphically to get the velocity and acceleration of the falling body. The cameras frame rate was not small enough to make calculations accurate therefore I used a method of interpolation of the graphs to make slope measuring accurate. The balls were dropped overall six times but only two drops were calculated later on in the essay since they were the only ones really straight and since the end graphs of both are very similar I assume that it is not needed to conduct the experiment more times as the end results would have very small differences.
The friction force versus time graphs got were close to theory and are explained by the formula for drag/friction force of water. The most interesting part of my results is that the terminal velocity got is similar to what would be expected to get using the formula of terminal velocity so overall I can say that the experiment proves that the air resistance theory is the same in water it only differs by the conditions of water which is that it is more dense and more viscous forces are acting on the ball during the fall than in air. This is probably the same for other mediums as well but should be investigate furthermore.
Word count: 274
Table of contents
1.Introduction
1.1.Equipment description
1.2.Procedure
2.Data collection, processing, presentation and analysis
2.1.Raw data table
2.2.Displacement versus time
2.3.Interpolated data tables and graphs
2.4.Velocity versus time
2.5.Acceleration versus time
2.6.Friction force versus velocity
2.7.Analysis
3.Conclusion and Evaluation
3.1.Conclusion
3.2.Evaluation
4.Bibliography:
- Introduction
At first I wanted to do an experiment involving air resistance like throwing different objects off the balcony and measuring the difference in falling. But to get reasonably distinguishable measurements I would need to drop these objects from a lot higher than a balcony which leads to my inability to do the experiment because I cannot get the heights needed. So I decided to do it in a medium in which the change in force and acceleration would be noticeable more quickly and in lower heights. This gave me an interesting idea to determine if the theory of air resistance applies in other mediums than air like water by simulating it and see if the simulations are accurate, according to the theory, or not and what is the reason for inaccuracies. So my research question is how force of resistance is dependent from objects velocity in water and whether the mechanics of air resistance are similar in water like in air.
1.1.Equipment description
The equipment I used was:
1)Glass vessel (long and thin so that dropped object has enough distance to fall vertically and its acceleration decrease noticeably but it does not have space to move horizontally)
2)Metal ball (for an object to drop, the shape of a ball was picked because I want it to move only vertically and that it does not get too strong resistance from the shape that it instantly is at the terminal velocity)
3)Video camera (In order to determine the velocity changes I decided to use a video camera and then work with the footage using program called VideoPoint because balls fell down the vessel very quickly and using something that can stop frames helps a lot)
4)Water
5)Pincers (Pincers were used to drop balls because the neck of the vessel was very thin to fit the ball with fingers)
1.2.Procedure
The experiment itself is not very complicated but it has a lot of uncertainties that involve the experiment itself and the calculations and graphical conversions of data. I filled the vessel with water and started dropping the balls while recording them with a stable camera. First ball used was very big and this lead to a lot of horizontal movement which was not good for accurate measurement. Then I decided to use smaller ones. All of the small balls were the same size for similar results and dropping them resulted in a straight fall. This might be compromised because the camera sees the ball in a 2D plane really. Ball is going from top to bottom and can move to left and to right according to camera but it can also move towards the camera and away from it which the camera would not capture it as the balls position in that plane is not visible. This movement can lead to small inaccuracies between the drops if one drop is completely straight and the other is moving in the plane that camera can see which I assume looking at the footage as straight. Sometimes it got a bit of a curve to left or right as well which might be because my hand shook while letting go. I dropped overall 6 balls then used the footage in VideoPointprogram to measure the time it took for the ball to fall a certain distance at different points in falling. This allows me to graphically calculate the velocity and acceleration of the ball at different points in the fall. Other inaccuracies can be for example the bending of light. If the ball drops not completely in the middle the bending of the glass vessel the glass will bend the light making the camera capture inaccurate movement of the ball. The inaccuracies in calculating come in as I am making a lot of conversions to graphs. Taking one data table’s information putting it into a graph then getting its slope making another graph from it which can result in greater inaccuracies if there was some errors in interpolation of functions. So the experiment has to be conducted very carefully because a small mistake at the start can make the end results vary a lot.
- Data collection, processing, presentation and analysis
Using the footage from the experiment I noticed that some drops were very curved while some were straight. So I only used three that were the most straight and the rest left as inaccurate.
2.1.Raw data table
Table 1.
First drop / Second drop / Third droptime,s ∆t=0.001 / displacement,m ∆s=0.005 / time,s ∆t=0.001 / displacement,m ∆s=0.005 / time,s ∆t=0.001 / displacement,m ∆s=0.005
0.04 / 0.006 / 0.04 / 0.005 / 0.04 / 0.081
0.08 / 0.030 / 0.08 / 0.026 / 0.08 / 0.157
0.12 / 0.069 / 0.12 / 0.058 / 0.12 / 0.222
0.16 / 0.104 / 0.16 / 0.099 / 0.16 / 0.288
0.20 / 0.153 / 0.20 / 0.148 / 0.20 / 0.350
0.24 / 0.211 / 0.24 / 0.201 / 0.24 / 0.416
0.28 / 0.272 / 0.28 / 0.260 / 0.28 / 0.483
0.32 / 0.335 / 0.32 / 0.319
0.36 / 0.396 / 0.36 / 0.383
0.40 / 0.458 / 0.40 / 0.448
0.44 / 0.506 / 0.44 / 0.508
Digital SONY camera was used. I looked up the specifications of the model of the camera, which was used in the experiment, from SONY page[1] and found out that shutter speed varies from 1/50s to 1/4000s so the time uncertainty during filming could be around 0.001s. From the video the uncertainty of displacement is somewhere 0.005m because I had to measure from the same point of the ball which is hard to do and can vary. I process data using graphical methods. And in order to get the change in velocity or acceleration I try to get the slope of a function of displacement versus time. (this is instantanious velocity which menas I need very small differences in time) My data is dependent on the frame speed of the camera which means my measurement is every 0.04 seconds which is not small enough. In order to get more data values at smaller intervals I interpolated the graph using curve fitting which lead to calculate values every 0.01 interval using this function:
2.2.Displacement versus time
Now using the data from the raw data table first of all I need to construct graphs which have the function of displacement versus time. Using these graphs I can evaluate whether the data got from the video point are useful and similar to theory. Also before I can use Graphical analysis to accurately measure the slope therefore measuring the instantaneous velocity I will need these graphs to get the function which I then will use for interpolation. The graphs constructed are shown in Graph 1. , Graph 2. and Graph 3. below.
Graph 1.
Graph 2.
Graph 3.
The third drop graph is a straight line which is not useful for me because I need the change of force against the change of velocity and in this graph its constant. So I decided not to interpolate the third drop as it gives no results. The first and second graph look similar and seem to be accurate to the theory. At first the balls have zero velocity as they are at rest. After they are dropped the the slope of the function is increasing because displacement depend from the speed and since earths gravity force is acting on them they have acceleration so the the velocity is inreasing. The rate at which slope is changing is decreasing and at around 0.3 seconds reaching terminal velocity. This means that acceleration of the balls is decreasing until it reaches zero which leads to an assumption that some other force (Friction force) is acting on it that is increasing as the balls velocity increases. When that force, because of speed, is equal to the gravity force the balls no longer have acceleration leaving them at terminal velocity.
2.3.Interpolated data tables and graphs
Now in order to evaluate how force of friction is dependant from the velocity I need to get the instantanious velocity at specific time intervals. This I decided to do graphically because the slope of the function of displacement versus time gives velocity . I use Graphical Analysis 3.1 program which allows me to manipulate graphics made from x and y cordinates points that I write in. This tool helps me to find the slope but before it can be used accurately I need to have more points. Instantanious velocity is gathered by calculating slope at very small time intervals and my raw data table has too big of the intervals. But the graph that I got from raw data table lets me determine the function of the grapth that I can interpolate. Using interpolation I can get as many points as I want at any time intervals. The interpolated data tables are shown in Table 2. and Table 3. and drawn graphs from that data are shown in are shown in Graph 4. and Graph 5. below. The uncertainties are the same as for raw data as the data varies by the same amount because this was used to get more points and the data was got from function.
Table 2.
First graph interpolationt,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005
0.00 / 0.000 / 0.11 / 0.053 / 0.22 / 0.184 / 0.33 / 0.350
0.01 / 0.001 / 0.12 / 0.063 / 0.23 / 0.198 / 0.34 / 0.365
0.02 / 0.002 / 0.13 / 0.073 / 0.24 / 0.213 / 0.35 / 0.381
0.03 / 0.005 / 0.14 / 0.083 / 0.25 / 0.228 / 0.36 / 0.396
0.04 / 0.008 / 0.15 / 0.094 / 0.26 / 0.243 / 0.37 / 0.411
0.05 / 0.012 / 0.16 / 0.106 / 0.27 / 0.258 / 0.38 / 0.426
0.06 / 0.017 / 0.17 / 0.118 / 0.28 / 0.273 / 0.39 / 0.440
0.07 / 0.023 / 0.18 / 0.130 / 0.29 / 0.288 / 0.40 / 0.455
0.08 / 0.029 / 0.19 / 0.143 / 0.30 / 0.304 / 0.41 / 0.469
0.09 / 0.037 / 0.20 / 0.156 / 0.31 / 0.319 / 0.42 / 0.483
0.10 / 0.045 / 0.21 / 0.170 / 0.32 / 0.335 / 0.43 / 0.496
0.44 / 0.509
Graph 4.
Table 3.
Second graph interpolationt,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005 / t,s ∆t=0.001 / s,m ∆s=0.005
0.02 / 0.001 / 0.13 / 0.068 / 0.24 / 0.201 / 0.35 / 0.368
0.03 / 0.003 / 0.14 / 0.077 / 0.25 / 0.215 / 0.36 / 0.384
0.04 / 0.007 / 0.15 / 0.088 / 0.26 / 0.230 / 0.37 / 0.400
0.05 / 0.011 / 0.16 / 0.099 / 0.27 / 0.245 / 0.38 / 0.416
0.06 / 0.015 / 0.17 / 0.110 / 0.28 / 0.260 / 0.39 / 0.432
0.07 / 0.021 / 0.18 / 0.122 / 0.29 / 0.275 / 0.40 / 0.447
0.08 / 0.027 / 0.19 / 0.134 / 0.30 / 0.290 / 0.41 / 0.463
0.09 / 0.034 / 0.20 / 0.147 / 0.31 / 0.306 / 0.42 / 0.478
0.10 / 0.041 / 0.21 / 0.160 / 0.32 / 0.321 / 0.43 / 0.494
0.11 / 0.050 / 0.22 / 0.173 / 0.33 / 0.337 / 0.44 / 0.509
0.12 / 0.058 / 0.23 / 0.187 / 0.34 / 0.353
Graph 5.
2.4.Velocity versus time
Now that I have many points at very small intervals I use the Graphic analysers tool to get the slope at specific time intervals which are the instantaneous velocity at those intervals the uncertainties now for those velocities are equal to sum of proportional uncertainties of time and displacement multiplied by the magnitude of velocity (calculated values displayed in Table 4. and Table 5.) and time uncertainty is the same. The data tables of instantaneous velocity of first and second drops are shown in Table 4. and Table 5. and the graphs of the instantaneous velocity of the balls created from those tables are shown in Graph 6. and Graph 7. below. Now the graphs that I expect to get from the instantaneous velocity should have the slope that is equal to instantaneous acceleration at specific time intervals as . The function should at first be increasing quickly and then decrease because acceleration at first should be close to 9.8 m/s2 and as the balls are reaching bottom that is when the balls speed is increasing the acceleration should reach zero. Interpolation for these graphs are no longer necessary since the graph is already created from the interpolated graph of displacement versus time which means graph will already have many points with small enough time intervals.
Table 4.
Instantaneous velocity of the first dropt,s / v,m/s / ∆v, m/s / t,s / v,m/s / ∆v, m/s / t,s / v,m/s / ∆v, m/s / t,s / v,m/s / ∆v, m/s
0.03 / 0.3 / 0.3 / 0.14 / 1.08 / 0.07 / 0.25 / 1.49 / 0.04 / 0.36 / 1.51 / 0.02
0.04 / 0.4 / 0.2 / 0.15 / 1.13 / 0.07 / 0.26 / 1.50 / 0.04 / 0.37 / 1.49 / 0.02
0.05 / 0.5 / 0.2 / 0.16 / 1.18 / 0.06 / 0.27 / 1.52 / 0.04 / 0.38 / 1.47 / 0.02
0.06 / 0.5 / 0.2 / 0.17 / 1.23 / 0.06 / 0.28 / 1.53 / 0.03 / 0.39 / 1.45 / 0.02
0.07 / 0.6 / 0.1 / 0.18 / 1.27 / 0.06 / 0.29 / 1.54 / 0.03 / 0.40 / 1.43 / 0.02
0.08 / 0.7 / 0.1 / 0.19 / 1.31 / 0.05 / 0.30 / 1.54 / 0.03 / 0.41 / 1.40 / 0.02
0.09 / 0.8 / 0.1 / 0.20 / 1.35 / 0.05 / 0.31 / 1.55 / 0.03 / 0.42 / 1.36 / 0.02
0.10 / 0.8 / 0.1 / 0.21 / 1.38 / 0.05 / 0.32 / 1.55 / 0.03 / 0.43 / 1.33 / 0.02
0.11 / 0.90 / 0.09 / 0.22 / 1.41 / 0.04 / 0.33 / 1.54 / 0.03 / 0.44 / 1.31 / 0.02
0.12 / 0.96 / 0.08 / 0.23 / 1.44 / 0.04 / 0.34 / 1.54 / 0.03
0.13 / 1.02 / 0.08 / 0.24 / 1.46 / 0.04 / 0.35 / 1.52 / 0.02
Graph 6.
Table 5.
Instantaneous velocity of the second dropt,s / v,m/s / ∆v, m/s / t,s / v,m/s / ∆v, m/s / t,s / v,m/s / ∆v, m/s
0.05 / 0.4 / 0.2 / 0.15 / 1.06 / 0.07 / 0.26 / 1.47 / 0.04
0.06 / 0.5 / 0.2 / 0.16 / 1.11 / 0.06 / 0.27 / 1.49 / 0.04
0.07 / 0.6 / 0.1 / 0.17 / 1.16 / 0.06 / 0.28 / 1.51 / 0.03
0.08 / 0.7 / 0.1 / 0.18 / 1.20 / 0.06 / 0.29 / 1.53 / 0.03
0.09 / 0.7 / 0.1 / 0.19 / 1.24 / 0.05 / 0.30 / 1.54 / 0.03
0.10 / 0.8 / 0.1 / 0.20 / 1.28 / 0.05 / 0.31 / 1.56 / 0.03
0.11 / 0.84 / 0.09 / 0.21 / 1.32 / 0.05 / 0.32 / 1.57 / 0.03
0.12 / 0.90 / 0.08 / 0.22 / 1.35 / 0.05 / 0.33 / 1.57 / 0.03
0.13 / 0.96 / 0.08 / 0.23 / 1.39 / 0.04 / 0.34 / 1.58 / 0.03
0.14 / 1.01 / 0.07 / 0.24 / 1.42 / 0.04 / 0.35 / 1.58 / 0.03
0.25 / 1.44 / 0.04 / 0.36 / 1.58 / 0.02
Graph 7.
I got very similar graphs to what I was expecting above. At first the balls’ speed was increasing quickly and as theygot closer to the ground acceleration slowly decreasedand finally reach zero at around 0.30 seconds as expected from the displacement versus time graphs because at that time the displacement was changing at a constant speed.
2.5.Acceleration versus time
As said above these graphs now allow me to get the instantaneous acceleration values of each drop in order to see how the acceleration is decreasing. This is also done using Graphical analyser and measuring the slope at every time interval. The data values of the instantaneous acceleration are shown in Table 6. below (uncertainties are calculated in the same way the velocity uncertainties were calculated. It is the sum of proportional uncertainties of time and velocity, which were calculated before, times the magnitude of acceleration at that time interval).
Table 6.
Instantaneous acceleration of the first dropt,s / a,m/s2 / ∆a m/s2 / t,s / a,m/s2 / ∆a m/s2 / t,s / a,m/s2 / ∆a m/s2
0.04 / 9 / 6 / 0.13 / 5.8 / 0.5 / 0.22 / 2.9 / 0.1
0.05 / 8 / 4 / 0.14 / 5.5 / 0.4 / 0.23 / 2.61 / 0.09
0.06 / 8 / 3 / 0.15 / 5.1 / 0.3 / 0.24 / 2.29 / 0.07
0.07 / 8 / 2 / 0.16 / 4.8 / 0.3 / 0.25 / 1.98 / 0.06
0.08 / 7 / 1 / 0.17 / 4.5 / 0.2 / 0.26 / 1.66 / 0.05
0.09 / 7 / 1 / 0.18 / 4.2 / 0.2 / 0.27 / 1.34 / 0.04
0.10 / 6.7 / 0.9 / 0.19 / 3.9 / 0.2 / 0.28 / 1.03 / 0.03
0.11 / 6.4 / 0.7 / 0.20 / 3.6 / 0.1 / 0.29 / 0.71 / 0.02
0.12 / 6.1 / 0.6 / 0.21 / 3.2 / 0.1 / 0.30 / 0.39 / 0.01
Instantaneous acceleration of the second drop
t,s / a,m/s2 / ∆a m/s2 / t,s / a,m/s2 / ∆a m/s2 / t,s / a,m/s2 / ∆a m/s2
0.06 / 7 / 3 / 0.16 / 4.8 / 0.3 / 0.26 / 2.38 / 0.07
0.07 / 7 / 2 / 0.17 / 4.5 / 0.3 / 0.27 / 2.12 / 0.06
0.08 / 7 / 1 / 0.18 / 4.3 / 0.2 / 0.28 / 1.87 / 0.05
0.09 / 6 / 1 / 0.19 / 4.1 / 0.2 / 0.29 / 1.63 / 0.04
0.10 / 6.2 / 0.9 / 0.20 / 3.8 / 0.2 / 0.30 / 1.39 / 0.03
0.11 / 6.0 / 0.7 / 0.21 / 3.6 / 0.1 / 0.31 / 1.15 / 0.03
0.12 / 5.8 / 0.6 / 0.22 / 3.3 / 0.1 / 0.32 / 0.90 / 0.02
0.13 / 5.5 / 0.5 / 0.23 / 3.1 / 0.1 / 0.33 / 0.66 / 0.01
0.14 / 5.3 / 0.4 / 0.24 / 2.84 / 0.09 / 0.34 / 0.418 / 0.008
0.15 / 5.0 / 0.4 / 0.25 / 2.60 / 0.08 / 0.35 / 0.176 / 0.003
As the tables show at start balls’ acceleration is around 7-9m/s2 which is close to free falling acceleration. As the ball was closer to the bottom of the glass vessel the acceleration got smaller until it reached zero and the balls were in terminal velocity.
2.6.Friction force versus velocity
From these tables I use the formula to get the Friction force and then construct a graph with function of Friction force versus velocity which show how Force of friction is dependent from velocity of each drop. The formula I use is: where (Ff) is Friction force, (m) is mass of the metal balls, (g) is free falling acceleration and (a) is the instantaneous acceleration of the metal ball. So using the table of instantaneous acceleration and inserting the acceleration into the formula and taking that balls mass was 0.075kg and both balls have the same mass I get the table of velocity and force which is shown in Table 7. below.
Table 7.
Instantanious velocity and Friction force for the first dropv,m/s / ∆v, m/s / Ff,N / ∆F, N / v,m/s / ∆v, m/s / Ff,N / ∆F, N / v,m/s / ∆v, m/s / Ff,N / ∆F, N
0.4 / 0.2 / 0.09 / 0.06 / 1.02 / 0.08 / 0.30 / 0.03 / 1.41 / 0.04 / 0.52 / 0.02
0.5 / 0.2 / 0.11 / 0.05 / 1.08 / 0.07 / 0.33 / 0.02 / 1.44 / 0.04 / 0.54 / 0.02
0.5 / 0.2 / 0.14 / 0.04 / 1.13 / 0.07 / 0.35 / 0.02 / 1.46 / 0.04 / 0.56 / 0.02
0.6 / 0.1 / 0.16 / 0.04 / 1.18 / 0.06 / 0.37 / 0.02 / 1.49 / 0.04 / 0.59 / 0.02
0.7 / 0.1 / 0.18 / 0.04 / 1.23 / 0.06 / 0.40 / 0.02 / 1.50 / 0.04 / 0.61 / 0.02
0.8 / 0.1 / 0.21 / 0.03 / 1.27 / 0.06 / 0.42 / 0.02 / 1.52 / 0.04 / 0.63 / 0.02
0.8 / 0.1 / 0.23 / 0.03 / 1.31 / 0.05 / 0.44 / 0.02 / 1.53 / 0.03 / 0.66 / 0.02
0.90 / 0.09 / 0.25 / 0.03 / 1.35 / 0.05 / 0.47 / 0.02 / 1.54 / 0.03 / 0.68 / 0.02
0.96 / 0.08 / 0.28 / 0.03 / 1.38 / 0.05 / 0.49 / 0.02 / 1.54 / 0.03 / 0.71 / 0.02
Instantanious velocity and Friction force for the second drop
v,m/s / ∆v, m/s / Ff,N / ∆F, N / v,m/s / ∆v, m/s / Ff,N / ∆F, N / v,m/s / ∆v, m/s / Ff,N / ∆F, N
0.5 / 0.2 / 0.19 / 0.07 / 1.11 / 0.06 / 0.38 / 0.02 / 1.47 / 0.04 / 0.56 / 0.02
0.6 / 0.1 / 0.21 / 0.06 / 1.16 / 0.06 / 0.39 / 0.02 / 1.49 / 0.04 / 0.58 / 0.02
0.7 / 0.1 / 0.23 / 0.05 / 1.20 / 0.06 / 0.41 / 0.02 / 1.51 / 0.03 / 0.59 / 0.02
0.7 / 0.1 / 0.25 / 0.04 / 1.24 / 0.05 / 0.43 / 0.02 / 1.53 / 0.03 / 0.61 / 0.02
0.8 / 0.1 / 0.27 / 0.04 / 1.28 / 0.05 / 0.45 / 0.02 / 1.54 / 0.03 / 0.63 / 0.02
0.84 / 0.09 / 0.29 / 0.03 / 1.32 / 0.05 / 0.47 / 0.02 / 1.56 / 0.03 / 0.65 / 0.01
0.90 / 0.08 / 0.30 / 0.03 / 1.35 / 0.05 / 0.49 / 0.02 / 1.57 / 0.03 / 0.67 / 0.01
0.96 / 0.08 / 0.32 / 0.03 / 1.39 / 0.04 / 0.50 / 0.02 / 1.57 / 0.03 / 0.69 / 0.01
1.01 / 0.07 / 0.34 / 0.03 / 1.42 / 0.04 / 0.52 / 0.02 / 1.58 / 0.03 / 0.70 / 0.01
1.06 / 0.07 / 0.36 / 0.03 / 1.44 / 0.04 / 0.54 / 0.02 / 1.58 / 0.03 / 0.72 / 0.01
Using this table I can construct the graph with function Friction force versus velocity which showshow Friction force is dependent from velocity. The problem with this is that I could not create a function that would allow me to get the Friction force graphically. This means my calculations for Friction force can have a big inaccuracy if any data error were entered in graphical calculations/measurements before or if there was any inaccuracies with the conducted experiment. The constructed graphs of Friction force versus velocity are shown in Graph 8. and Graph 9. below.
Graph 8.
Graph 9.
The two graphs above are very similar to each other. At first velocity is increasing quickly as is the friction force but as balls fall the speed at which velocity changes (acceleration) is decreasing. The speed of change of friction force at the end is decreasing slower than velocity therefore it bends upwards. At the end of the graph the graph cuts out because terminal velocity is reached. At this point velocity does not change any more just like friction force so the graph cannot go further beyond that point. Since the two graphs are very similar it seems that more drops were not necessary as the result graph would not give any particular difference that could be analysed. So the two drops are sufficient for determining the dependency of friction force versus time.
2.7.Analysis
The Table 7. is now the most important table as it holds my research question’s answer of how is friction force dependant from velocity. At the start it is expected that there is no friction force exerted on the ball from Table 7. it is shown that the friction force is a little bigger than zero that is 0.08886N for the first drop and0.19458N for the second. This is because the calculations are not exactly at the start. The starts of the graphs are thrown away as it needs two additional points to calculate the slope and for the first point overall only two points are available and to make it even more accurate I left out two points from the start. This was also done for the ends of the graphs. And since the slope was calculated not only for one but for two different graphs the start and end cut off two times leaving our last graph starting from around 0.04 seconds of the fall instead of the start. Having that in mind the start friction forces are small enough to assume that at start they were zero. Now the maximum friction force that can be exerted on the balls cannot exceed the weight of the ball that is around 0.75 N (The balls mass is 0.075kg.0.075 * 9.8m/s2). The friction force at the end of the falls are 0.70545N and 0.7218N which are close to the expected weight of the balls. Friction forces at start and end of the falls in air should be the same so trying to compare these to some data gathered from research is not leading to better understanding of how the friction force is different in water. Now terminal velocity on other hand could give a lot of information. At the end of the fall where the balls reach terminal velocity speeds are 1.544593 m/s and 1.582196 m/s considering that they are not exactly reached terminal velocity 1.6 m/s is a close approximation of the true terminal velocity in my experiment. Terminal velocity is dependant from four variables the weight of the body, drag coefficient, density of the gas, and area of the object (, V – Terminal velocity, W – weight of the body, Cd – drag coefficient, – density of the gas, A – Frontal area)[2](„The drag coefficient is a number that aerodynamicists use to model all of the complex dependencies of shape, inclination, and flowconditions on aircraft drag“[3]). These determinants of the terminal velocity were given for the gasses. If in our case we take water, what determinants of terminal velocity change: the weight is the same, frontal area is the same. The only changes are the density of the medium, which is a lot bigger than air, and Coefficient of the drag. The drag coefficient is changed as it is dependent not only from the shape of the object but from the flow of the object in the medium in other words it depends from the viscosity of the medium. In water viscous forces should be stronger than in air because the surface of the object is being acted by a bigger amount atoms of water therefore I assume that drag coefficient should be bigger as well. A small golf ball weighs around 0.046 g which is a little bit less than weight of my ball and in air its terminal velocity is around 32 m/s.[4]Now assuming that my ball‘s terminal velocity in air is of the golf ball‘s that is 32 m/s the ratio is . That is in water ball has 20 times slower terminal velocity. Now assuming if it was only dependant from the density ( – water density, – air density).[5] If the only difference was density terminal velocity should be 28 time slower which knowing that because of the drag coefficient difference would be even bigger makes our 20 time slower terminal velocity seem inaccurate. This is of course because I took terminal velocity of a golf ball. My ball was heavier and is smoother and has smaller frontal area than the golf ball therefore meaning that terminal velocity of metal ball in air would be greater than 32 m/s leaving with the assumption that the 28 time slower or more is plausible as of my ball in reality is bigger than 20. Also any more points that I could have gotten from the experiment would have been put at the same point in the graph when terminal velocity was reached as the velocity and friction force would not change past that point. Thus in the graphs near that point where terminal velocity is reached that bending of function complicates my finding of a function that could define the relation of velocity and friction force, so I decided to only take the beginning of the graph, that is v ≈ 0.5m/s, till near the end, v ≈ 1.41m/s. At this part the relation seems to be close to quadratic function which seems viable as drag function in fluids is depending on the square of velocity (As seen in Graph 10. and Graph 11. below.)(Formula for drag force in fluids - ; - the drag force, which is by definition the force component in the direction of the flow velocity, - the mass density of the fluid, - is the flow velocity relative to the object, - is the drag coefficient – a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag, - is the reference area)[6].