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Importance of angle of release on a 3-point shot in basketball
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Rationale
My dad was a sports person till his college days, not like the casual ones we play. He professionally played basketball for years. Due to certain unavoidable circumstances, he could not pursue his career in basketball and had to join the family business.
He left sports but the spirit of sportsmanship never left him. Needless to say, how disciplined he was, he taught me the same. He encouraged me for sports from a very young age. He would take me out to grounds to run, exercise and play. He had his work pressure but never failed to manage time to take me out for play.
As I started growing older, my friends stopped coming to play regularly. They preferred video games but my dad still continued taking me to grounds and playing along.
It was my 14th birthday when my uncle gifted a home basketball set to me. I was happy beyond words could express. I grew up hearing incidents from my father's glorious events in basketball and dreamt of walking in his shoes.
That basketball became my life. My dad helped me a lot and I gained immense interest in it.
When I was in 11th grade, I changed my school. To my amazement, the school had a huge basketball court. Till then I had just played at home with dad or occasionally, with a few friends of mine. Playing with a complete team in such a big court was something else. I loved it anyway. I joined the school's team and started training. With the help and the guidance of the school coach and my father, my started playing quite well.
A sport is not just to learn how to play on the field. It has a lot of theory as well. One needs to study a lot as well. One such interesting fact that I had a real tough time in grasping was the 3-point shot.
This IA is about the same. It is to find out how the release angle of the basketball on a 3-point shot determines the success percentage of a professional player.
Aim
The main motive of this IA is to study the importance and range of release angle for a successful scoring in Basketball. Moreover, the motion of basketball after it is thrown is a projectile motion. So, another importance of this IA is to study the different parameters of the projectile or the Basketball that affects the pathway for a successful basketball score.
Research question
How does the release angle of basketball on a 3-point shot in basketball determine the shot percentage of professional players?
Background information
The trajectory of a Basketball is a projectile. Projectile is a motion in which the body propagates in both X- direction and Y- direction. But, considering the simplicity of the calculations, I will calculate the motion in both the dimensions separately.
As this IA deals with projectile motion, we will use a certain concept of Physics of Projectile Motion or Motion in a Plane. The parameter equations of motion in one direction are as follows:
- Displacement – Time relationship:
s = ut + \(\frac{1}{2}at^2\)
- Velocity – Time relationship:
v = u + at
- Velocity – Displacement relationship:
v2 = u2 + 2as
In the above-mentioned equations, s is displacement, u is initial velocity, v is final velocity after time t and a is the acceleration of the body.
As, the projectile is a motion in both directions, we will use these equations separately for both the dimensions. Such as:
For Propagation in X – Direction:
SX = uX t + \(\frac{1}{2}\)aXt2
vX = ux + aXt
\(v^2_X = u^2_X+2a_XS_X\)
We know that, there is no acceleration in X direction. Therefore, aX = 0. Therefore, we can write,
SX = uX t……(1)
vX = ux……(2)
For Propagation in Y – Direction:
SY = uY t \(\frac{1}{2}\)aYt2
vY = uY + aYt
\(v^2_Y = u^2_Y+2a_YS_Y\)
We know that, there is no acceleration in Y direction. Therefore, aY = - g = - 9.81 m/s2.
Therefore, we can write,
SY = uY t - \(\frac{1}{2}\)gt2……(3)
vY = uY - at……(4)
Apart from Mathematical formulae that has been derived from physics, there are few more pre-requisites of this IA. As this IA deals with scoring in Basketball game, some information about the basketball court is very essential.
Radius of 3-point score line in NBA: 7.24 m
Radius of 3-point score line in FIBA: 6.75 m
Diameter of Basket: 0.45 m
Height of Basket: 3.05 m
Average velocity of required to throw the ball for 3-point score: 10 m/sec.
For solving quadratic equation, Sreedhar Acharya’s Formula will be used in this IA, which is also shown below. If a quadratic equation is in the form of:
ax2 + bx + c = 0
Then the value of x will be equal to:
\(x=\frac{-b±\sqrt{b^2-4ac}}{2a}\)
Introduction
In this IA, we will be finding the value of release angle for successfully scoring of a 3-point shot. Using the concepts of projectile motion, I will find the trajectory of the basketball, hence, I will conclude with the exact values of angle of release required. As there are more than one parameter responsible for the trajectory of a projectile, I will assume a value of the velocity of throwing a basketball for 3-point shot from a Basketball survey website. With the values of the angle that will be deduced below, I will explain the effect of uncertainty of the angle of release in trajectory concluding with a percentage of success of achieving proper basketball shot for different height of players.
Process of calculation
From the diagram it is evident that, the distance the ball will travel in horizontal axis will be equal to 7.24 m in NBA and 6.75 m in FIBA. In this IA, we will mainly focus on the NBA guidelines. The distance the ball will travel in vertical axis will vary with the height of the player. If the height of the player be h metres, then the displacement of the ball in vertical axis will be (3.05 – h) metres.
From equation (1), (2), (3) and (4), we can write,
For motion in X direction:
SX = 7. 24
uX = 10 × cos cos θ
Therefore, we can write,
SX = uXt
=> 7.24 = 10 cos cos θ ×t
=> t × cos cos = 0. 724 sec
=> t = 0. 724 sec sec θ
For motion in Y direction:
SY = 3. 05 - h
As, with different height of players, the angle of release will be different, we will consider few heights which are more often in basketball players for simplicity of calculation.
h = 1.62 metres
sY = 3. 05 - 1. 62 = 1. 43
\(s_Y=u_Yt+\frac{1}{2}a_Yt^2\)
=> 1. 43 = 10 sin sin θ t - \(\frac{1}{2}\)×9.81×t2
=> 1. 43 = 10 sin sin × 0. 724 sec sec θ - \(\frac{1}{2}\)× 9. 81 × (0. 724 sec sec θ )2
=> 2. 56 sec2 θ - 7. 24 tan tan θ + 1. 43 = 0
=> 2. 56 (1+ ) - 7. 24 tan tan θ + 1. 43 = 0
=> 2. 56 -7. 24 tan tan θ + 1. 43 + 2. 56 = 0
=> 2. 56 -7. 24 tan tan θ + 3. 99 = 0
Using Sreedhar Acharya’s Formula, we get,
tan tan θ = \(\frac{7.24±\sqrt{(-7.24)^2-4×2.56×3.99}}{2×2.56}\)
tan tan θ = 1. 41 ± 0. 66
∴ θ1 = 64. 2° ± 0. 01° , θ2 = 36. 8° ± 0. 1°
h = 1.72 metres
sY = 3. 05 - 1. 72 = 1. 33
\(s_Y=u_Yt+\frac{1}{2}a_Yt^2\)
=> 1. 33 = 10 sin sin θ ×t - \(\frac{1}{2}\) × 9. 81 × t2
=> 1. 33 = 10 sin sin θ × 0. 724 sec sec θ - \(\frac{1}{2}\)× 9. 81 × (0. 724 sec sec θ )2
=> 2. 56 sec sec 2θ - 7. 24 tan tan θ + 1. 33 = 0
=> 2. 56 (1 + θ) - 7. 24 tan tan θ + 1. 33 = 0
=> 2. 56 -7. 24 tan tan θ + 1. 33 + 2. 56 = 0
=> 2. 56 - 7. 24 tan tan θ + 3. 89 = 0
Using Sreedhar Acharya’s Formula, we get,
tan tan θ = \(\frac{7.24±\sqrt{(-7.24)^2-4×2.56×3.89}}{2×2.56}\)
tan tan θ = 1. 41 ± 0. 69
∴ θ1 = 64. 5° ± 0. 01° , θ2 = 35. 7° ± 0. 1°
h = 1.82 metres
sY = 3. 05 - 1. 82 = 1. 23
sY = uY t + \(\frac{1}{2}\)aYt2
=> 1. 23 = 10 sin sin θ ×t - \(\frac{1}{2}\)× 9. 81× t2
=> 1. 23 = 10 sin sin θ × 0. 724 sec sec θ - \(\frac{1}{2}\)× 9. 81 × (0. 724 sec sec θ )2
=> 2.56 sec2 θ - 7. 24 tan tan θ + 1. 23 = 0
=> 2. 56 (1 + θ ) - 7. 24 tan tan θ + 1. 23 = 0
=> 2. 56θ - 7. 24 tan tan θ + 1. 23 + 2. 56 = 0
=> 2. 56θ - 7. 24 tan tan θ + 3. 79 = 0
Using Sreedhar Acharya’s Formula, we get,
tan tan θ = \(\frac{7.24±\sqrt{(-7.24)^2-4×2.56×3.79}}{2×2.56}\)
tan tan θ = 1. 41 ± 0. 72
∴ θ1 = 64. 8° ± 0. 01° ,θ2 = 34. 6° ± 0. 1°
h = 1.92 metres
sY = 3. 05 - 1. 92 = 1. 13
sY = uY t + \(\frac{1}{2}\)aYt2
=> 1. 13 = 10 sin sin θ ×t - \(\frac{1}{2}\)× 9. 81×t2
=> 1. 13 = 10 sin sin θ ×0. 724 sec sec θ - \(\frac{1}{2}\)× 9. 81 × (0. 724 sec sec θ )2
=> 2. 56 sec2 θ - 7. 24 tan tan θ + 1. 13 = 0
=> 2. 56 (1 + θ) - 7. 24 tan tan θ + 1. 13 = 0
=> 2. 56θ - 7. 24 tan tan θ + 1. 13 + 2. 56 = 0
=> 2. 56 - 7. 24 tan tan θ + 3. 69 = 0
Using Sreedhar Acharya’s Formula, we get,
tan tan θ = \(\frac{7.24±\sqrt{(-7.24)^2-4×2.56×3.69}}{2×2.56}\)
tan tan θ = 1.41 ± 0.75
∴ θ1 = 65. 1° ± 0. 01°, θ2 = 33. 4° ± 0.1°
h = 2.02 metres
sY = 3. 05 - 2. 02 = 1. 03
sY = uY t + \(\frac{1}{2}\)aYt2
=> 1. 03 = 10 sin sin θ ×t - \(\frac{1}{2}\)× 9.81×t2
=> 1. 03 = 10 sin sin θ × 0. 724 sec sec θ - \(\frac{1}{2}\)× 9. 81 × (0. 724sec θ )2
=> 2. 56 sec2 θ - 7. 24 tan tan θ + 1. 03 = 0
=> 2. 56 (1 + θ ) - 7. 24 tan tan θ + 1.03 = 0
=> 2. 56θ - 7. 24 tan tan θ + 1.03 + 2. 56 = 0
=> 2. 56 - 7. 24 tan tan θ + 3. 59 = 0
Using Sreedhar Acharya’s Formula, we get,
tan tan θ = \(\frac{7.24±\sqrt{(-7.24)^2-4×2.56×3.69}}{2×2.56}\)
tan tan θ =1. 41 ± 0. 78
∴ θ1 = 65. 4° ± 0. 01°, θ2 = 32. 2° ± 0. 1°
In this IA, I have shown this graph to develop a trendline for the players with height other than the figures shown in the above section. The equation of the trendline that should be followed by various height of players to score 3-point shot with the proper angles considering the initial velocity of the ball is 10 m/sec is shown below:
y = - 11. 379x + 55. 228
Another graph is also there for the other values of theta or the angle of release with respect to height of players:
y = 3x + 59. 34
Successful shot percentage
From the above study we can get, there are two types of uncertainty based on the value of angle of release. For higher values of , the maximum percent error or uncertainty is equal to:
\(\frac{0.01}{64.2}\)× 100 = 0. 0155 %
Similarly, the minimum percent error or uncertainty is equal to:
\(\frac{0.01}{65.4}\)× 100 = 0. 0152 %
So, we can conclude that, for higher values of θ or the angle of release, the percentage uncertainty is 0.15%. Thus, the percentage of successful shot with the release angles as mentioned above with initial velocity of 10 m/s is 99.985%.
For lower values of θ, the maximum percent error or uncertainty is equal to:
\(\frac{0.01}{32.2}\)× 100 = 0. 310 %
Similarly, the minimum percent error or uncertainty is equal to:
\(\frac{0.01}{36.8}\)× 100 = 0. 271 %
So, we can conclude that, for higher values of θ or the angle of release, the percentage uncertainty is 0.29%. Thus, the percentage of successful shot with the release angles as mentioned above with initial velocity of 10 m/s is 99.708%.
Conclusion
In this IA, I have studied the importance of angle of release of basketball for a 3-point score. As the motion of basketball is a projectile so we have used the projectile equations of motion and concluded that, with definite angles of release, a successful 3-point score can be achieved considering the initial velocity of the ball just after throw is 10 m/sec. Though with change in velocity, the angle of release for successful shot will change, but in this IA, the velocity is considered to be constant. The velocity of throwing a basketball for 3-point shot is usually 10m/sec as preferred by the players. As we know that, with increase in height of basketball player, the ball should be thrown such that the displacement in Y axis decreases. As a result, with increase in height, the angle of release of basketball will also change. According to the above work, I can conclude that, considering the initial velocity to be constant as 10m/s, for successful scoring of 3-points, the angle of release should be 64.2° or 36.8° for a player with height of 5 feet 4 inches, 64.5° or 35.7° for a player with height of 5 feet 8 inches, 64.8° or 34.6° for a player with height of 6 feet, 65.1° or 33.4° for a player with height of 6 feet 4 inches and 65.4° or 32.2° for a player with height of 6 feet 8 inches. To find the angle of release for successful shot for players with heights other than the mentioned values, the equation of trendline is shown as: y = -11.379x + 55.228 or y = 3x + 59.34. Despite of a wide basket of diameter 0.45 metres, with a minute change in angle, the trajectory will change to such an extent that the ball will not go through the basket. In order to determine the successful shot percentage, for initial velocity being 10 m/sec, the results ranges between 99.708% to 99.985% depending upon the angle of release.
Bibliography
- Miller, Stuart, and Roger Bartlett. "The relationship between basketball shooting kinematics, distance and playing position." Journal of sports sciences 14.3 (1996): 243-253.
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