When the ball reached the highest point the vertical velocity was 0.
Vertical:
v = 0 m/s
a = -9.8 m/s2
u = 7.42 m/s
a = (v – u) / t
t = (v – u) / a = 0.76 s
Horizontal:
u = 12.58 m/s
a = 0 m/s2
t = 0.76×2 = 1.52 s
S = ut + 1/2 at2 = 12.58 ×1.52= 19.12 m
That is my rough prediction based on rough data
Max height = 2.81 m
Range = 19.12 m
Analysing
Near Shot
Height:
First of all I will investigate the height of the shots. I have plotted the graph which showed above. It’s a Distance - Time Graph. From the graph you can see clearly that the height of the ball increases but if you can compare first four points of the graph you will find that it increases less because of gravity. The gravity makes the ball decelerate. And the vertical velocity goes down. Then the height increases less.
Now I am going to calculate the highest point of the near shot. I will use the method previously in the rough calculation but more explanations.
Initial vertical velocity =the gradient of the first point =distance change between first two points divides time change between first two points.
“u” ─ initial vertical velocity “H1” ─ height of the first point
“H2” ─ height of the second point “t1” ─ time of the first point.
“t2” ─ time of the second point
u = (H2 – H1) / (t2 – t1) = (0.38 – 0.0) / (0.04 – 0.00) = 9.50 m/s
I am going to use the formula “v2= u2 + 2as” because we know when the ball reaches the highest point the vertical velocity of the ball is 0 and the initial velocity is 9.50m/s. Also we know the magnitude of acceleration of the gravity is 9.8 m/s2.When we have already known the values of these three then we can calculate the distance.
v2= u2 + 2as
2as= v2 - u 2
s= (v2 - u 2) / 2a = (02 – 9.502)/(2×(-9.80)) = 4.60cm=4.60m
You should notice that the acceleration in the formula is negative because the acceleration is a vector. It’s got the direction. Assuming that it is positive value when the direction goes down. Here the motion of the ball is upward then it’s got a negative value.
Range
Secondly I want to calculate the range of the near shot. Form the graph you can see the distance in an interval time does not change too much, because the vertical acceleration of the ball is 0 if ignore the air resistance. Without the acceleration the velocity will not change at all. In the real life you can’t ignore the air resistance so the rate of increasing the distance is slight changed.
First I should work out the initial velocity horizontally. Basically the method is the same as calculating the initial vertical velocity.
“u” ─ initial horizontal velocity “D1” ─ distance of the first point
“D2” ─ distance of the second point “t1” ─ time of the first point.
“t2” ─ time of the second point
u = (D2 – D1)/ (t2-t1) = (0.54 – 0.00)/ (0.04-0.00) = 13.50m/s
The time for the ball to reach the highest point is the same as to move the half range so if I can work out how long it takes to reach the highest point then I can calculate the range.
Vertical: a = (v-u)/t
t = (v – u)/a = (0 – 9.50) / (-9.80) = 0.97s
Horizontal:
s = ut = 13.50 × 0.95 = 13.10cm = 13.10m
So the whole range = 13.10 × 2 = 26.20 m
Here you should separate the calculation into two parts vertical and horizontal.
Do not be confused by using the same symbols for the different calculation.
Initial velocity
The initial horizontal and vertical velocity can be calculated by drawing the graph.
It will be shown on the graph paper.
Initial vertical velocity = 9.29 m/s
Initial horizontal velocity = 15.28 m/s
Initial velocity = √ (Initial vertical velocity)2 + ( Initial horizontal velocity)2
= √ (9.29) 2 + (15.28) 2
= √ (319.78)
= 17.88 m/s
I have plotted a velocity against time graph for all the points. The blue dots present the horizontal velocity. The violet dots present the vertical velocity. Both of them trend to be negative. The deceleration make the velocity go down. Horizontally the deceleration is air resistance. Vertically the deceleration is gravity.
Far Shot:
I will use the method which I have shown previously.
The highest point was calculated.
The height = 5.96m
Compare with the value on the data it is a little bit different. Because in my calculation I assume that the acceleration is 9.8 but in really it can not be exactly 9.8. This is why it is different.
The Range was calculated. (Method was shown previously)
The range = 31.63m
Also here there is different between the values on the data. For the reason there is air resistance so it is quite different.
Velocities are calculated by drawing the graphs.
Initial horizontal velocity = 17.86 m/s
Initial vertical velocity = 11.11 m/s
Initial velocity was calculated by using the vectors
Initial velocity = 21.03 m/s
Comparison of results
All the results above are based on parabolic equations. The method that I used to calculate the near shot and far shot are exact the same but I think far shot is more accurate because it contains more data. The precision errors are less. Look at all the data of the far shot and the near shot. The data of the far shot are all bigger than the near shot. There are only 7slides for the near shot. It is not very accurate. Also the air resistance will affect the ball.
Precision error
In both near shot and far shot the camera is placed in line with the centre of the part of the ball’s path. This means that the first and last point further from the camera than in the middle points.
The near shot was scaled from the tennis ball launcher, which is at the edge of the photograph. Therefore, the scaling was correct for the edge to the photographs, but there should be some systematic drift towards the middle of the photograph. As the camera was close to the ball it would make the ball’s velocity larger than the true value.
In the initial horizontal and vertical calculation there are some random errors in drawing and reading.
The uncertainty of taking the value of the time is+/-0.002s
There are some errors on the basic data. I have labeled on the graph.
Data
