Data Analysis - Two cameras were set up to record images of a tennis ball launcher firing a tennis ball into the sky.
Data Analysis Coursework
Introduction
Two cameras were set up to record images of a tennis ball launcher firing a tennis ball into the sky.
Camera 1 was set up in line with the camera launcher and filmed the beginning of the launch.
The film was then digitised and 7 sides were produced of the ball's motion. Each slide is separated by 0.04 seconds (these slides are numbered from 1 to 7). Camera 2 was set up in line with what was thought to be the halfway point of the ball's motion. This camera filmed the whole of the ball's motion from the start to the end of it. The film was then digitised into 25 slides of photos. Each slide is separated by 0.08 seconds (these slides are numbered from 1 to 45 in odd numbers).
Planning
I was given an excel file fill with some measurements from the physics department. My teacher showed me the slides, which are printed out. After I had a chance to see it I recognize that they were taken a few years ago and it took place at the tennis court outside the physics department. The measurements that I have got were the near shot displacement and the far shot displacement. They measure by measuring the horizontal and vertical distance of the tennis ball from the tennis launcher. Calculations of the ratio of the near shot and for shot to the real life objects were made (i.e. The door of the physics department). My aim is to work out some of the answer below with the data I've decided to use tables and graphs, in order to present the data more clearly. Graphs may be able to show some relationship between variables.
From the near shot photos: The angle when the ball was launch
How fast the ball is travelling
The Horizontal Distance and speed
The Vertical Distance and speed
From the far shot photos: How high the ball went
The angle when the ball is landing
The horizontal distance and speed when falling
The vertical distance and speed when falling
Background Information
Photo
Time/s
Y
0.00
0.00
2
0.04
5.10
3
0.08
37.80
4
0.12
64.20
5
0.16
94.40
6
0.20
22.80
7
0.24
43.60
In this experiment when the ball launches from the tennis ball launcher it travels towards the sky, due to the launcher was aimed to the sky. The ball starts to decelerate as soon as it is launched due to air resistance, when the ball reaches it's maximum height the ball will start to accelerate down towards the ground as the gravity on earth pulls it down. If we got the measurements, we can use this formula S= ut+1/2at² of SUVAT which can calculate the height and range that the ball went. The results are then used to plot a graph in order to show more relationships between the data. I predict there will be a smooth curve. But due the formula does not include any air resistance so the graph will be a bit higher then reality.
The prediction of the graph that I expect
Near shot
These are the measurements of the near shot. And the X and Y represent the horizontal and vertical distance at different time. The data also includes the scaled up measurements, the distance in real life. It was calculated by using the given ratio, which was 1:20 so times the each values of x and y by 20 in the photo to the vertical and horizontal distance in real life.
Near Shot
Displacement/cm
Photo
Scaled up
Photo
Time/s
x
...
This is a preview of the whole essay
The prediction of the graph that I expect
Near shot
These are the measurements of the near shot. And the X and Y represent the horizontal and vertical distance at different time. The data also includes the scaled up measurements, the distance in real life. It was calculated by using the given ratio, which was 1:20 so times the each values of x and y by 20 in the photo to the vertical and horizontal distance in real life.
Near Shot
Displacement/cm
Photo
Scaled up
Photo
Time/s
x
y
x
Y
0.00
0.00
0.00
0.00
0.00
2
0.04
2.70
.90
54.0
38.0
3
0.08
5.50
3.50
10
70.0
4
0.12
8.00
5.00
60
00
5
0.16
0.80
6.90
216
38
6
0.20
2.50
7.50
250
50
7
0.24
5.10
8.90
302
78
Analysis of the Data
This is the graph of vertical distance (Y) against time.
This is a distance-time graph so the gradient of the graph is the velocity. A curve is shown on the graph which means that the velocity is changing. After the ball is fired, it is set up to fire in an angle, which shoots towards the sky. As soon as the ball is fired it is decelerating downwards as you can see from the graph then generally it slows down even more at (0.24, 178) because of gravity is pulling it towards the ground and the air resistance increases. At the beginning it is curving up this means that it is under deceleration but then it curves down. This means that the ball has reaches its maximum height and starts to drop towards the ground (Falling down). From this graph, also you can see my prediction was proved to be right because I had mentioned earlier that the ball will travel towards the sky.
By finding out the initial horizontal velocity and the initial vertical velocity we can then calculate the maximum height and how far did the ball travelled, by using the SUVAT equation S=ut+1/2at² to solve it. (S=Displacement, U=Initial velocity, T= Time, A= Acceleration (Gravity), V= Finial Velocity)
We can find out both of the horizontal and vertical initial velocity in the way shown below:
54 / 100 to get centimetres into meters
= 0.54 m
Then 0.54 / 0.04 to get the initial horizontal velocity
= 13.5 m/s
38 / 100 to get centimetres into meters
= 0.38
Then 0.38 / 0.04 to get the initial vertical velocity
=9.5 m/s
Why doing this would helps in the S=ut+1/2at² equation, this is because we have got only a little bit of the values in the equation. As displacement we can find them out from the results tables, the acceleration that is the gravity for the vertical must be 9.8m/s, for the horizontal it must be -9.8. Final velocities for both axes are 0 as it lands on the floor etc. So to complete the equation we have to find out some other information that we can find out from these results such as the initial velocity for both axes.
By this stage we have got both the initial vertical velocity and initial horizontal velocity therefore we can plug in them into the equation with other measurements that we already know in order to work out the range of the ball. (UX=13.5m/s, UY=9.5m/s)
As we already know that the final vertical velocity of the ball is 0m/s, we also know that the acceleration of the ball vertically is -9.8 as gravity is 9.8 m/s at this latitude so it is decelerating at 9.8 m/s. Plus there is no horizontal acceleration as there are no force acting upon as we are ignoring the air resistance.
Here is how I am going to use these values in order to find out other results which I want by using the SUVAT equations:
Vertical
S= 4.8 U= 9.5 m/s V= 0 m/s A= 9.8 m/s T= 0.96s
S= ut+1/2at²
S= 13.5 x 0.96 + (0.5 x 0 x 0.96²)
S= 12.96 x 2
2S= 25.92 m
Maximum range is 25.92m
Horizontal
S= 25.92 U= 13.5 m/s V=? m/s A= 0 m/s T= 0.96s
V= u + at
v- u /a= t
9.5/9.8= 0.96s
S= ut+1/2at²
S= 9.5 x 0.96 + (0.5 x -9.8 x 0.96²)
S= 4.6m
Maximum height is 4.6m
From these measurements, I have found out that the initial vertical displacement is 25.92m/s and the horizontal vertical displacement is 4.6m/s. To make sure that my measurements and equation are right I can look up the other piece of data from the physics department, they are the far shot photos (shown below).
By comparing data of these photos, the maximum range that the tennis ball travelled was 25.76m, and from my calculations I got 25.92m, which was very close. Despite there was a bit of error in the horizontal distance, I have got the vertical distance in the exact value to the data.
This graph is plotted from the far shot result. From this graph we can see the speed of the ball.
The graph above shows a curve which means that the velocity is changing. Also we can see the gradient of the graph is equal to the acceleration. At the start after the ball is fired, it speeds up towards 600m/s as the gradient is curving up which means that it is under deceleration but then it is curving down (slows down) as it reaches its maximum height at 524cm/s, as gravity pulls it down (meaning that there is gravity acting on the tennis ball)
To find out the velocity going up and down we can calculate as below:
Distance/time
Going upwards Going downwards:
57/100 91/100
=0.57m =0.91m
0.57/0.08 0.91/0.08
=7.125m/s =11.4m/s
As you can see the velocity going downwards is 11.4m/s, it's really close to the gravitational force (approximately 10m/s)
Far Shot
Displacement/cm
Photo
Scaled up
Photo
Time/s
X
y
X
y
0.00
0.20
0.10
22.8
1.4
3
0.08
.50
0.90
71
03
5
0.16
2.70
.80
308
205
7
0.24
4.00
2.50
456
285
9
0.32
7.20
3.00
821
342
1
0.40
6.40
3.50
730
399
3
0.48
7.50
3.90
855
445
5
0.56
8.50
4.30
969
490
7
0.64
9.60
4.50
094
513
9
0.72
0.6
4.60
208
524
21
0.80
1.6
4.50
322
513
23
0.88
2.7
4.40
448
502
25
0.96
3.7
4.20
562
479
27
.04
4.7
3.90
676
445
29
.12
5.8
3.60
801
410
31
.20
6.6
3.20
892
557
33
.28
7.5
2.90
995
331
35
.36
8.5
2.40
2109
274
37
.44
9.3
.70
2200
94
39
.52
20.1
0.90
2291
03
41
.60
21.0
0.40
2394
45.6
43
.68
21.7
-0.70
2474
-79.8
45
.76
22.6
-1.30
2576
-148
Far shot
From the graph shown the page below we can see there are some errors in the data, there is a plot slightly off the curve at the beginning and the middle. The two points are located at where the tennis ball is climbing and falling. I carefully read the data. I tried to understand why these errors had occurred. After reading the data carefully I found out that there was a calculation problem and a measurement error in the data.
The first error is from photo 9 and it is the fifth point on the graph. I carefully remeasure the photo again I found out that the actual horizontal distance was 5.2cm instead of 7.2cm, so after multiplied it by the ratio which is 114, I got the scaled up as the new X value is 593. The second error that I found was a calculation of the scaled up Y value in photo 31 instead of multiple the result by 114 they multiple it with 174 and got 557 instead of 365. After I corrected it, the curve perfectly fitted.
Graph of far shots
Graph of far shots with correction in data
Errors
Due the errors in this experiment, I've produced a graph with errors bars in it. This means that the curve passes through the error bars. These errors are probably link to air resistance, measurement errors, gravity and the position where the photo was taken.
Conclusions
In this experiment I was given a set of data, which was record a few years ago of a tennis ball travel from a tennis ball launcher, and I was told to investigate with it.
I used the near shot photo's data with the SUVAT equation as S= ut+1/2at²in order to find out how far the tennis ball went (travelled) and how high it goes.
The result of what I found was very close to the actual data (far shot) by checking with the other set of data I can prove myself that my equation was correct, the reason I think the courses of error is due to the fact that we are told to ignore air resistance so the error occurs, therefore I got the horizontal distance further.
Vertical
S= 4.8 U= 9.5 m/s V= 0 m/s A= 9.8 m/s T= 0.96s
S= ut+1/2at²
S= 13.5 x 0.96 + (0.5 x 0 x 0.96²)
S= 12.96 x 2
S= 25.92 m
Maximum range is 25.92m (2592cm) Results from data is 2576cm (scaled up X)
Horizontal
S= 25.92 U= 13.5 m/s V=? m/s A= 0 m/s T= 0.96s
V= u + at
v- u /a= t
9.5/9.8= 0.96s
S= ut+1/2at²
S= 9.5 x 0.96 + (0.5 x -9.8 x 0.96²)
S= 4.60416m
Maximum height is 4.6m (2.d.p) (460cm ±0.5cm) Results from data is 524cm(scaled up Y)
I believe that if we are not ignoring air resistance, the vertical distance shouldn't be that close with the data, so I think there was a bit of error with both of the distances, but overall the error was accepted.
Expect air resistance, other factors that will effect the overall measurements are gravity, the errors when measuring the distance with the photos, and the angle of the camera when it is recording the movement of the ball. Because if the camera wasn't placed perpendicular with the line that the ball travelled, it will make the measurements go wrong and affects the results of the experiment.
Because I didn't actually saw the experiment when it was taken place so I'm not sure about this problem. But overall the calculations and the results are very close together this proves that the method of which I used is right.
Page 1 12/18/2007
