Task: Find values for the initial speed and angle of projection needed to score a basket in basketball from the free throw line. Is there a range of values of angle for an initial velocity that will still allow you to score a basket?

Method: To solve this problem I am going to record myself scoring 10 clean baskets (the ball does not touch the hoop or back board) on video camera. Having done this I will know the time taken from the ball leaving my hands to it going into the basket. By using vector equations for Displacement, Velocity and Aceeleration in relation to time I will be able to calculate:

* The maximum height achieved by the ball.

* The distance travelled by the ball

* The time taken to travel this distance

* The height of release and the height at which it lands.

This will give me an accurate model of the balls motion through the air. I will be able to use the velocity vector equations to find the horizontal and vertical components of velocity for each throw. Using Pythagoras Theorem I will be able to calculate the initial velocity. Using trigonometry I will also be able to calculate the angle of release. This will allow me to analyse how initial velocity and the angle of release are related and therefore allow me to find a range of values for the angles of release for a given velocity.

Assumptions and Facts: As this is a simplified model there are a few things which have been assumed. Firstly I have assumed the ball to be a particle at the balls centre of gravity. This simplifies the fact that the displacement of the ball would be different for each side of it so it wouldn’t fit the equation.

Secondly I have ignored the effect of any air resistance which would normally cause friction and slow the ball down thus changing its motion. This is because air resistance would be very difficult to calculate as it changes depending on the velocity of the ball. Also as I am conducting the throws on an outdoor court the air resistance could change depending on the wind.

Finally I have assumed that when the ball goes cleanly through the basket it goes directly through the centre of the basket. This allows me to keep the horizontal length of all throws the same as the ball starts and lands in the same places.

From research I have found the relevant dimensions of the courts. Firstly I know the distance from the free throw line to the middle of the basket (the horizontal distance travelled by the particle) to be 4.572m. Secondly I know the height of the basket to be 3.048m.

Notation

I will take gravity to be 9.81N represented by g

Velocity will be Represented by V and measured in m/s

Time will be represented by t measured in seconds

Displacement will be represented by r in metres

Results and Calculations: From research and data collection I know the distance from the start to the finish of the motion, the time taken for the ball to reach the basket and the release height. I know that the acceleration of the of the particle is represented by [ ]

Integrating this allows me to find the vector equation for the velocity of the particle. Although I don’t know the initial conditions of the motion yet they are represented by V and V so the equation becomes [ ]

Again integrating this allows me to find a vector equation for the position of the particle. For simplicity I have modelled the situation so that the origin of the axis is point of release of the ball hence the initial conditions of the position are (0,0) so the equation is [ ]

I have used calculations to 3 decimal places because the constants I have put in (e.g. the horizontal length) are all to 3 decimal places and therefore this is the only degree to which I am sure I can be accurate.

I have found the release height to be 1.168m. This enabled me to find the vertical displacement of the basketball as it went through the basket: Height of Basket – Height of Release =3.048 -1.88=1.168

Below is an example of a calculation:

To work out the initial velocity of the ball I used the velocity vector equation. For the horizontal velocity component we have the equation V x t = r. We know that when t is 1.01 seconds r is 4.572m. This can be rearranged to allow us to find V:

Another rearrangement can be done for the vertical component of velocity as we know that after 1.01 seconds r is 1.168 m:

Using the vertical and horizontal components of velocity I can now use Pythagoras theorem to find the initial velocity:

Using trigonometry I know that the angle of release; Sin ? = Opposite / Hypotenuse

To find the maximum height achieved by the ball I again use the velocity vector equation. I know that when the ball is at its maximum height its vertical velocity is 0 m/s therefore : – 9.81t + V = 0

Because I already know the value of V I can rearrange to find the time at which the Vertical velocity is equal to zero.

I can now substitute this value of t into the vertical component of position vector equation to give me the height of the ball at this time:

This table shows information on the motion of each throw.

Throw No.

Time (seconds)

Maximum Height (metres)

Angle of Release

(degrees)

Initial Velocity

(metre per second)

1

1.01

1.903

53.468

7.605

2

1.12

2.178

58.014

7.706

3

1.03

1.950

54.339

7.614

4

1.01

1.903

53.468

7.605

5

1.1

2.125

57.232

7.679

6

1.07

2.049

56.021

7.645

7

1.16

2.286

59.521

7.770

8

1.12

2.178

58.014

7.706

9

1.06

2.024

55.608

7.636

10

1.09

2.099

56.833

7.667

Average

1.077

2.070

56.252

7.663

I have also investigated the velocity needed to achieve a basket for any given angle. Although I dont know the values for the horizontal and vertical components of velocity they can be represented by

[ ]

These values can be put into the position vector equation to give:

[ ]

I know that the horizontal component is equal to 4.572 m so I can rearrange to give:

This can then be substituted into the vertical component of the position equation. This leaves only t as an unknown quantity so:

This gives the total time for the throw. This value can now be substituted into the velocity vector equation to find the initial velocity as shown earlier.

I have calculated the initial velocity of a number of angles, starting at 14.4 as this is the minimum angle for which it would be possible to score a basket.

Angle (degrees)

Initial Velocity (metres per second)

Maximum Height (metres)

14.4

25.306

2.589

20

15.210

1.396

30

9.638

1.184

40

8.092

1.379

50

7.614

1.734

60

7.794

2.322

70

8.771

3.462

80

11.718

6.788

89

35.935

65.798

So far I have assumed that the ball goes into the centre of the basket each time, but it is obviously possible for the ball to land at either end of the basket and still go in. If the basket is 45cm in diameter then the ball can go through the basket at any point between 4.347m and 4.797m away from the thrower. This means that for a constant initial velocity there is a range of angles for which the ball would still go in. In order to find this range I used a decimal search method. I know that at a certain time the ball is 4.347m away I also know that at this point it is also 1.168m high. By trial and error I was able to find the time at which the horizontal vector equation was equal to 4.347m and the vertical vector equation was equal to 1.168m. Consequently I was able to find the angle of release for the ball.

I have plotted a number of graphs: Angle vs Velocity; Height vs Angle, Height vs Velocity these will make analysis of my results easier and also allow me to find values for variables which I haven’t calculated.

Interpreting and Validating: Having plotted my results in graphs they have shown me a number of things. Firstly, the graph off release angle against initial velocity for my shots shows that the greater the angle of release the greater the initial velocity needed to get it to the basket. For example the throw with the least angle of 53.648 degrees had the least velocity at 7.605m/s whilst the throw with the greatest angle 59.521 degrees had the greatest velocity of 7.770m/s. This suggests that the maximum angle for throwing a free throw could be anything up to 90 degrees (as the ball would go straight up.). For example if the release angle of the ball was 89degrees you could still score a basket if you could give the ball a large enough initial velocity. Giving the ball a large enough velocity would be a problem as I calculated that to score a basket with that release angle you would have to give it a velocity of 35.935m/s. The graph I plotted to show this does not follow a linear relationship and it shows that for a given change in angle the velocity increases by a greater amount the higher the angle. The velocity needed for a ball released at 89 degrees is roughly 5 times the speed at which I scored my baskets. This is obviously an improbable speed to achieve and so for a human the maximum angle would be less – possibly around 70 – 75 degrees where the velocity needed would be a much more achievable 8 – 9 m/s. Although this is an attainable velocity for the ball it means that you have to apply a greater force to the ball, which consequently would reduce the accuracy of your shot, and therefore you would have less chance of scoring a basket.

When looking at the minimum angle for scoring a basket I have found that the angle is likely to be about 30 degrees. Although the minimum angle with which you could get to the basket is 14.4 degrees this would not allow you to score a basket as the ball reaches its maximum height after passing the basket hence it is still travelling up when it gets to the basket and therefore would not be able to go through the basket. This means that when looking for the minimum angle the ball has to reach its maximum height before the basket so that it can fall through it. From using the minimum turning point on my graph of maximum Height against release angle it is possible to estimate the minimum angle more accurately to 31 degrees. But again scoring a basket at this angle would require a large velocity of 25.306m/s.

The ball has the best chance of going through the basket when it is falling at an angle near vertical, as there is more margin for error. However we have seen that the large angles require the ball to have too much velocity. The most favourable angle would be one with a high trajectory but also one that requires a comfortable throwing velocity so that the ball can be thrown accurately. Again using the turning point of my graph showing velocity against angle I can see that this angle is roughly 54 degrees.

I also found that for an initial velocity of 7.770m/s there was a range of angles between 55 degrees and 62.352 degrees which still allowed a basket to be scored.

In general my conclusions were:

* A basket can be scored at any angle between 31 degrees and 89 degrees providing you are able to give it a great enough velocity.

* The most favourable angle was roughly 54 degrees as it had a high trajectory and didn’t require a high velocity.

* For a constant initial velocity there is a large range of angles for which you can still score a basket.

Problems: The first problem is that the range of values that I calculated for the release angle would be dramatically decreased if we considered the whole ball as opposed to a particle at its centre of gravity. For example I calculated that the ball could land at either end of the hoop however this would not actually be possible because if the Centre of gravity of the ball landed in the edge of the basket the rest of the ball would hit the rim meaning you wouldn’t score a basket. I found the diameter of a basketball to be approximately 25cm. This means that the centre of the ball cant land within 12.5cm of the rim if a basket is to be scored. Hence the centre of the ball has to land within a 20 cm diameter circle of the centre of the basket.

Secondly I said that in theory the ball could be thrown at 89 degrees and if it was given enough velocity it would still go in the basket. In reality this is unlikely because the greater the force you give to the particle the greater the air resistance and friction which consequently change the motion of the ball.

The fact that I assumed that when I scored the ball had gone through the exact centre of the basket means that a number of my calculations could be inaccurate because as shown the ball can go in the basket without going through the exact centre. This means that when I used the horizontal distance travelled by the ball as 4.572m (the distance to the centre of the hoop) it wasn’t necessarily the actual distance travelled by the ball. This means that all the calculations made from it would be incorrect.

Finally, the experiment in general had the possibility for a lot of errors. For example there was human error involved in the timing of each throw and the measuring of the release point so the answers derived from these values would be inaccurate. Also because there are so many calculations it means that any error gets increasing multiplied to cause an even bigger inaccuracy. Having said this there isn’t a lot that can be done to gain more accurate results without using advanced equipment. For example light gates could be used to measure the time of the throw more accurately and machines could be used to throw the ball so that the release height, velocity and angle could all be better controlled.

There are a number of things that could be investigated to achieve a more accurate representation of the angles needed to score a basket. Firstly I only looked at the angle of release for one release height. If the release height was higher or lower then the release angle could change or the velocity needed for a certain angle would be different which might make the optimum angle for different people. Secondly you could look at the affect of jump shots from the free throw line, again this is increasing the release height.