Mathematical modeling of the projectile motion of volleyball and determining the equation of its trajectory

Mathematics is a very versatile and essential subject to analytically understand any real life scenario. It is extensively used in classical mechanics which deals with the motion of objects. I have always been a volleyball enthusiast and I wanted to know the physics and the math behind it. I wanted to understand the motion of the volleyball from a mathematical perspective but lacked the faintest idea of how to apply mathematics in such a real life scenario.

 

In my higher studies, I got introduced to motion in a plane and there I learnt about projectile motion and how effectively mathematics is used to analyze the motion of an object (in air). It gave me the slightest idea of how mathematics could be used to describe the motion of volleyball in air and gave me the mathematical perspective of imagining things which I have been longing for. I have tried explaining mathematically the motion of a volleyball and its trajectory in this project with whatever expertise I have gained in mathematical modeling from projectile motion and motion of an object in a plane in general.

 

The study would help an enthusiast like me to get an idea of how mathematics is extensively used to study the motion of an object in air and its trajectory (here, the motion and trajectory of volleyball). It would also help to get an idea of how a mathematical model is necessary to analyze such real life scenarios and to optimize the serve of volleyball.

Mathematical modeling of the projectile motion of volleyball and determination of the equation of its trajectory.

Projectile is the name given to a body thrown with some initial velocity with the horizontal direction, and then allowed to move in two dimensions under the action of gravity alone, without being propelled by any engine or fuel. This motion of the projectile is called projectile motion. The path followed by a projectile is called its trajectory. To study the motion of a projectile, we will consider the following:

• There is no air resistance
• The effect due to the rotation of the Earth and curvature of the Earth is negligible
• The acceleration due to gravity is constant in magnitude and direction at all points of the motion of the projectile

Vectors are physical quantities which have both magnitude and direction for example;

 

Velocity, acceleration, force etc.

Let two vectors \(\overrightarrow A\) and \(\overrightarrow B\) act at some angle to each other as shown in Fig. (a). Coinciding the tail of the vector \(\overrightarrow A\)  on the head of vector \(\overrightarrow B\), we get the resultant vector \(\overrightarrow R\) as shown in Fig. (b).

 

The vectors \(\overrightarrow A\) and \(\overrightarrow B\) are represented by the sides \(\overrightarrow {OP}\) and \(\overrightarrow {PQ}\)  of the triangle OPQ.

 

Vector addition and resolution of vectors can be used to resolve a force vector into its vertical and horizontal components.

In the above figure, \(\overrightarrow F\) is the force vector which makes an angle θ with respect to the x-axis. This force \(\overrightarrow F\) can be resolved into its two rectangular components using basic trigonometry. The vertical component is F sin sin θ and the horizontal component is F cos cos θ as shown in the above figure.

A Cartesian coordinate system is a coordinate system in a plane that assigns each point an ordered pair of numerical coordinates which are the distances to the point from the two fixed axes, which are perpendicular to each other. The point of intersection of these two axes is called the origin whose position is characterized by the ordered pair (0, 0).

The study has been done mainly in three steps. Firstly, the projectile motion of the volleyball from its point of serve to where it lands has been considered, analyzed and the equations of motion has been derived. Secondly, the equation of the trajectory of the volleyball has been derived using the equations of motion and expressed in terms of the coordinates of the points through which the volleyball passes. Lastly, some numerical examples have been done to get a better understanding of the parameters involved in the motion of the volleyball.

In the above diagram,

 

The volleyball has been considered a point object and,

 

H is the height of the volleyball net

 

h_max is the maximum height attained by the volleyball

 

h₀ is the initial height of the volleyball at the location of serving

 

V is the initial velocity with which the volleyball is served

 

θ is the initial angle the volleyball makes with the horizontal at the serve location

 

Point A is the initial position of the volleyball at the serve location

 

Point B is the location, just above the net, through which the volleyball passes to the other side of the court

 

Point C is the location where the volleyball lands

 

g is acceleration due to gravity

 

L₁ is the distance between the net and the player who serves the volleyball from point A

 

L₂ is the distance the final position C of the volleyball and the net

To derive the equations of projectile motion of the volleyball at a position (x, y), we have from the equations of motion, x = (V cos cos θ)t…(i) where V cos cos θ is the component of the initial serve velocity of the volleyball along the horizontal or x-axis and t is the time (in seconds).

 

y = (V sin sin θ)t - \(\frac{1}2\) gt²…(ii) where V cos cos θ is the component of the initial serve velocity of the volleyball along the vertical or y-axis, g is acceleration due to gravity and t is the time (in seconds).

 

From equation (i) we get,

 

x = (V cos cos θ) t

 

t = \(\frac{x}{V\ cos\ cos\ θ}\)

 

Now putting this value of t in equation (ii) we get,

 

y = (V cos cos θ) t - \(\frac{1}{2}\)gt²

 

y = (V cos cos θ) \(\frac{x}{V\ cos\ cos\ θ}-\frac{1}{2}g(\frac{x}{V\ cos\ cos\ θ})^2\)

 

y = x tan tan θ - \(\frac{gx^2}{2(V\ cos\ cos\ θ)^2}\)...(iii)

 

The obtained equation (iii) represents the equation of a parabola in terms of x and y. The general form of a parabola whose axis is parallel to the y-axis is,

 

y = ax² + bx + c

 

By comparing this standard equation of a parabola with equation (iii) we get,

 

a = - \(\frac{g}{2(V\ cos\ cos\ θ)^2}\)

 

b = tan tan θ

 

c = 0

 

The coordinates of the point B and C will satisfy the equation of the trajectory of the volleyball i.e., equation (iii).

 

Coordinates of point B relative to the coordinate system xy (with origin at point A) is (L₁, mod(H - h₀)) For our convenience, we assume that the height of the net H is greater than the height from which the volleyball is served.

 

∴ Coordinates of point B relative to the coordinate system xy (with origin at point A) is (L₁, mod(H - h₀)

 

Coordinates of point C relative to the coordinate system xy (with origin at point A) is (L₁, mod(H - h₀)

 

Now replacing x and y by the coordinates of the point B in equation (iii) we get,

 

y = x tan tan θ -\(\frac{gx^2}{2(v\ cos\ cos\ θ)^2}\)

 

H - h₀ = tan L₁ tan - θ \(\frac{gL_1^2}{2(v\ cos\ cos\ θ)^2}\)

 

2(V cos cos θ)² (H - h₀) = 2L₁ (V cos cos θ )² tan tan θ - gL₁²

 

2(V cos cos θ)² (H - h₀) = 2L₁ V²(cos cos θ )² \(\frac{sin\ sin\ θ}{cos\ cos\ θ}\) - gL₁²

 

2(V cos cos θ)² (H - h₀) = L₁ V² cos cos θ  sin sin θ - gL₁²

 

∴ gL₁² - L₁ V² sin sin 2θ + 2 (V cos cos θ )²  (H - h₀) = 0...(iv)

 

Now replacing x and y by the coordinates of the point C in equation (iii) we get,

 

y = x tan tan θ - \(\frac{gx^2}{2(V\ cos\ cos\ θ)^2}\)

 

- h₀  = (L₁ + L₂) tan tan - θ - \(\frac{g(L_1+L_2)^2}{2(V\ cos\ cos\ θ)^2}\)

 

- 2h₀(V cos cos θ )² = 2(V cos cos θ )² (L₁ + L₂) tan tan θ - g(L₁ + L₂)²

 

g(L₁ + L₂)² - 2 (V cos cos θ)² (L₁ + L₂) tan tan - 2h₀(V cos cos θ )² = 0…(v)

 

From the obtained equations (iv) and (v), we can calculate L₁ and L₂ if all the other parameters other than the constant or fixed ones are known. Similarly, other parameters can be calculated.

In the above diagram,

 

The volleyball has been considered a point object and,

 

Point A is the initial position of the volleyball at the serve location

 

Point B is the location, just above the net, through which the volleyball passes to the other side of the court

 

Point C is the location where the volleyball lands

 

L₁ is the distance between the net and the player who serves the volleyball from point A

 

L₂ is the distance the final position C of the volleyball and the net

 

d is the distance between the final position or location of the volleyball and the net

 

α is the angle which the trajectory of the volleyball makes with the side line

 

l m is the length and breadth (in meters) of the volleyball court

 

From diagram 3, we observe that,

 

L₁ cos cos a = l

 

And,

 

L₂ cos cos α = d

 

Therefore,

 

L₁ = \(\frac{l}{cos\ cos\ α}\)

 

L₂ = \(\frac{d}{cos\ cos\ α}\)

 

We have the equation of the parabola

 

y = x tan tan θ - \(\frac{gx^2}{2(V\ cos\ cos\ θ)^2}\)...(iii)

 

And the two equations derived by replacing the values of x and y by the coordinates of point B and point C

 

gL₁² - L₁V ² sin sin 2θ + 2 (V cos cos θ )² (H - h₀) = 0…(iv)

 

g (L₁ + L₂)² - 2(V cos cos θ)² (L₁ + L₂) tan tan θ - 2h₀ (V cos cos θ)² = 0...(v)

 

Now we will put the values of L₁ = \(\frac{l}{cos\ cos\ α}\) and \(\frac{d}{cos\ cos\ α}\) in these equations to find the expressions in terms of d, l and α.

 

gL₁² - L₁V ² sin sin 2θ + 2 (V cos cos θ )² (H - h₀) = 0…(iv)

 

g\((\frac{l}{cos\ cos\ α})^2-\frac{l}{cos\ cos\ α}\)V ² sin sin 2θ + 2(V cos cos θ)² (H - h₀) = 0

 

gl² - V ² cos cos α sin sin 2θ + 2(H - h₀) (V cos cos θ cos cos α )² = 0…(vi)

 

g(L₁ + L₂)² - 2 (V cos cos θ)² (L₁ + L₂) tan tan θ - 2h₀ (V cos cos θ )² = 0…(v)

 

g\((\frac{l}{cos\ cos\ α}+\frac{d}{cos\ cos\ α})^2\) - 2(V cos cos θ )²\((\frac{l}{cos\ cos\ α}+\frac{d}{cos\ cos\ α})\)tan tan θ - 2h₀ (V cos cos θ )² = 0

 

g\((\frac{l+d}{cos\ cos\ α})^2\) - 2(V cos cos θ )²\((\frac{l+d}{cos\ cos\ α})\frac{sin\ sin\ θ}{cos\ cos\ θ}\) - 2h₀ (V cos cos θ )² = 0

 

g(l + d)² - 2 sin sin θ cos cos θ V ² (l + d) cos cos α - 2h₀ (V cos cos θ cos cos α )² = 0

 

g(l + d)² - V ² (l + d) sin sin 2θ cos cos α - 2h₀(V cos cos θ cos cos α )²  = 0…(vii)

 

Now the time for which the volleyball remains in air is,

 

t = \(\frac{L_1+L_2}{V\ cos\ cos\ \theta}\)

 

t = \(\frac{\frac{l}{cos\ cos\ α}+\frac{d}{cos\ cos\ α}}{V\ cos\ cos\ θ}\)

 

t = \(\frac{l+d}{V\ cos\ cos\ θ\ cos\ cos\ α}\)

 

Now, to get an idea for how long the volleyball remains in air, let us take l = 9m, d = 6m, initial serve velocity of the volleyball V = 3m/s, the initial angle which the volleyball makes with the horizontal at the serve location θ = 15°, the angle which the trajectory of the volleyball makes with the sideline α = 30°.

 

Therefore, the time spent by the volleyball in air is,

 

t = \(\frac{l+d}{V\ cos\ cos\ θ\ cos\ cos\ α}\)

 

t = \(\frac{9+6}{V\ cos\ cos\ 15°\ cos\ cos\ 30°}\) secs

 

t = \(\frac{15}{3×0.965×0.866}\)secs

 

t = 5.98 secs

 

Actually, the angle α does not change the time the volleyball spends in air but affects the horizontal component of the initial serve velocity or the horizontal speed V cos cos θ of the volleyball. This implies that the greater the angle α, the greater the speed of the volleyball.

 

Now, for given values of initial serve velocity V, height of the net H, initial serve height h₀, the angle α which the trajectory of the volleyball makes with the sideline, the initial angle θ which the volleyball makes with the horizontal at the serve location, we will calculate the length l using equation (vi).

 

So, let V = 2.8m/s, H = 5m, h₀ = 2m, α = 40°, θ = 17°. Now we will find the value of l  by these values in equation (vi).

 

gl² - V ² cos cos α sin sin 2θ + 2(H - h₀)(V cos cos θ cos cos α)² = 0…(vi)

 

9.8 × l² - (2.8)² cos cos 40° sin sin (2 × 17°) + 2(5-2)(2.8 × cos cos 17° cos cos 40°)² = 0

 

9.8 × l² - 7.84 × 0.766 × 0.559 + 2 × 3 × (2.8 × 0.956 × 0.766)² = 0

 

9.8 × l² - 3.357 + 6 × 2.05² = 0

 

9.8 × l² - 3.357 + 25.215 = 0

 

9.8 × l² = - 21.858

 

l² = \(\frac{21.858}{9.8}\) …[we ignore the negative sign since, it has no significance here]

 

l = \(\sqrt{2.23}m\)

 

∴ l = 1.49m

 

Now by putting this value of l and the value of the other parameters which we have assumed earlier in equation (vii) to find the value of distance d between the final position or location of the volleyball and the net.

 

g(l + d)² - V ² (l + d) sin sin 2θ cos cos α - 2h₀ (V cos cos θ cos cos)² = 0…(vii)

 

9.8 × (1.49 + d)² - 2.8²(1.49 + d) sin sin (2 × 17)° cos cos 40° - 2 × 2(2.8 × cos cos 17° cos cos 40°)² = 0

 

9.8 × (1.49 + d)² - 7.84 × (1.49 + d) sin sin 34° cos cos 40° - 4 × 2.05² = 0

 

9.8 ×(1.49 + d)² - 3.358 × (1.49 + d) - 16.81 = 0

 

9.8y² - 3.358y - 16.81 = 0…[Let y = (1.49 + d)]

 

y = \(\frac{3.358±\sqrt{(-3.358)^2-4×9.8×(-16.81)}}{2×9.8}\)

 

y = \(\frac{3.358±\sqrt{11.276+658.952}}{2×9.8}\)

 

y = \(\frac{3.358±\sqrt{670.228}}{19.6}\)

 

y = \(\frac{3.358±25.88}{19.6}\)

 

y = 1.4917 or y = - 1.149

 

Since, the negative sign has no significance here.

 

∴y = 1.4917

 

∴1.49 + d = 1.4917

 

∴d = 0.017m

 

Therefore, the value of the distance between the final position or location of the volleyball and the net d is 0.017m. This value depends on the initial height, initial serve velocity and the angle which it makes at the location of serve with the horizontal.

 

Similarly, we will find out the initial height h₀ of the volleyball at the location of serving using (vii).

 

So, let length and breadth of the volleyball l = 2m,

 

The distance between the final position or location of the volleyball and the net d = 1m,

 

The angle which the trajectory of the volleyball makes with the side line α = 30°,

 

The initial angle which the volleyball makes with the horizontal at the serve location θ = 20°,

 

Initial serve velocity of the volleyball V = 2.9m/s

 

Now, putting these values of the parameters in equation (vii) we get,

 

g(2 + 1)² - V ² (2 + 1) sin sin 2θ cos cos α - 2h₀ (V cos cos θ cos cos α)² = 0…(vii)

 

9.8 (2 + 1)² - 2.9² (2 + 1) sin sin (2 × 20)° cos cos 30° - 2h₀(2.9 cos cos 20° cos cos 30° )² = 0

 

9.8 × 3² - 2.9² × 3 × 0.642 × 0.866 - 2h₀(2.9 × 0.939 × 0.866)² = 0

 

88.2 - 14.027 - 11.12h₀ = 0

 

11.12h₀ = 74.173

 

h₀ = \(\frac{74.173}{11.12}\)

 

∴ h₀ = 6.67m

 

Therefore, the value of the initial serve height h₀ is 6.67m. Although, the serve height is an independent parameter but, can be calculated if the other parameters are known.

According to our analysis made and the scenarios, parameters considered which are involved in the projectile motion of the volleyball, the following conclusions can be made:

• The time which the volleyball spends in air depends upon the length L₁ between the point of serve and the net; the length L₂ between the net and the final position of the volleyball; the initial serve velocity and the serve angle.
• To reduce the time spend in air, the length L₂ should be made as large as possible since, the volleyball will reach a lower maximum height thereby, spending less time in air.
• If the initial height from which the ball is served is made as large as possible, the volleyball will reach its maximum height in air sooner thereby, spending less time in air.
• The value of the distance between the final position or location of the volleyball and the net d depends on the initial height, location of serve, initial serve velocity and the angle which it makes at the location of serve with the horizontal.
• The trajectory of the volleyball represents a parabola. The equation of the parabola passes through the points (L₁, H - h₀) and (L₁ + L₂, - h₀). So, these points satisfy the equation and from those equation unknown parameters can be determined.

• The motion of the volleyball in air has been analyzed from two different perspectives (side view and top view of the trajectory), which makes the study coherent.
• The general equations of motion of the volleyball and the general equation of its trajectory have been determined, which makes the study reliable to use in a real scenario to determine and optimize certain parameters.
• Numerical example have been done using the equations derived to get a better understanding of the dependency of the parameters and which also give us an understanding how the equations can be used for real life purposes.

• Certain important real life physical factors such as air resistance, aerodynamic forces acting on the volleyball, spin of the volleyball etc. have not been considered in the study.
• The volleyball has been considered a point object. The motion of the volleyball and the interdependence and the relation between the forces acting on the ball and the motion of the ball could be analyzed if the motion of the volleyball was studied in a three dimensional plane, considering the volleyball to be three dimensional object having mass.
• Not studying the forces acting on the volleyball and the relation between these forces and the motion of the volleyball makes the study less coherent and reliable.

Mathematics is essential to analyze various aspects and the motion of an object in air to make important conclusions and thereby, understand the physics behind the motion and the trajectory traced by the object. Also to understand the relation between the various parameters related to the motion and directly or indirectly affecting the motion, a mathematical analysis is needed. The above study is done excluding certain real life factors related to the motion of volleyball in air such as resistance, aerodynamic forces acting on the volleyball, spin of the volleyball etc. As an extension to this study, one could also analyze these real life factors, how they influence and directly or indirectly affect the motion of volleyball in air. Another study could be framed to study the relation between these real life factors and the parameters which have studied to analyze the motion of the volleyball. Apart from the extrinsic factors, intrinsic factors such as the mass of the volleyball could be considered and studied (for example, the relation between the mass of the volleyball and the distance between the final position or location of the volleyball and the net and derivation of an equation).

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