1. Cardinal Wuerl North Catholic High School
In Ace You Did Not Know
Alex Dudash
Cardinal Wuerl North Catholic High School
Grade 10
4th Year Participant

2. Introduction Serving is an integral component of volleyball.
I wanted to explore the physics of a good serve.
In this simulation:
a serve with a starting velocity of 7 meters per second
follows a direct, straight path
include the effect of air resistance, but not spin.
Constants used in calculation:
Average male height
Official net height
Official court dimensions

3. Introduction
I judged the serve’s success by observing the ball’s height at 9m (distance to net) and 18m (distance to end line), for angles 1⁰ -90⁰.
If the ball’s height at 9m is greater than the difference between the initial height and the net height ( m), it clears the net.
If the ball’s height at 18m is less than or equal to the floor’s height relative to the starting point (-1.77m), then it is within the court’s confines.
If both requirements are met the serve is successful.

4. Inspiration
I was influenced to do this project by my passion for volleyball. It is one of my high school activities, and I have played it since fifth grade.
Like many other sports, volleyball revolves around its fundamentals. Proper execution of serving, in particular, is crucial because it introduces every point.
I wanted to observe the physics behind a volleyball serve, and see the precision necessary to be a consistent server.

5. Development
Net ( m)  
9 meters
1.77 m
18 meters

6. Development
To calculate the ball’s motion, I first used the equation for drag force
𝐹 𝑑 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 2
𝐹 𝑑 represents the drag force
𝜌 represents the density of the air
𝐶 𝑑 represents the drag coefficient
𝐴 represents the cross-sectional area of the ball
𝑉 represents the velocity

7. Development
𝐹 𝑑 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 2
The equation for drag force

8. Development The equation for drag force
𝐹 𝑑 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 2
The equation for drag force
Assuming that the only forces acting upon the ball are gravity and air resistance, we know that
𝐹=−𝑚×𝑔− 𝐹 𝑑

9. Development The equation for drag force
𝐹 𝑑 = 1 2 ×𝑝× 𝐶 𝑑 ×𝐴× 𝑉 2
The equation for drag force
Assuming that the only forces acting upon the ball are gravity and air resistance, we know that
We also know that in the 2nd Law of Newton
𝐹=−𝑚×𝑔− 𝐹 𝑑
𝐹=𝑚× 𝑑𝑉 𝑑𝑡

10. Development The equation for drag force
𝐹 𝑑 = 1 2 ×𝑝× 𝐶 𝑑 ×𝐴× 𝑉 2
The equation for drag force
Assuming that the only forces acting upon the ball are gravity and air resistance, we know that
We also know that in the 2nd Law of Newton
Using the transitive property, these equations can combine to form
𝐹=−𝑚×𝑔− 𝐹 𝑑
𝐹=𝑚× 𝑑𝑉 𝑑𝑡
−𝑚×𝑔− 𝐹 𝑑 =𝑚× 𝑑𝑉 𝑑𝑡

11. Development
The force of drag can be divided by its vertical and horizontal components, and then inserted in the equation −𝑚×𝑔− 𝐹 𝑑 =𝑚× 𝑑𝑉 𝑑𝑡
− 𝐹 𝑑𝑥 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 2 × cos 𝜃
− 𝐹 𝑑𝑥 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉× 𝑉 𝑥
1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉×𝑉𝑥=𝑚× 𝑑 𝑉 𝑥 𝑑𝑡
− 𝐹 𝑑𝑦 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 2 × sin 𝜃
− 𝐹 𝑑𝑦 = 1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉× 𝑉 𝑦
−𝑚×𝑔− 1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉× 𝑉 𝑦 =𝑚× 𝑑 𝑉 𝑦 𝑑𝑡

12. Development
It is important to note that gravity is only included in the vertical equation, because it is a purely vertical force.
1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉×𝑉𝑥=𝑚× 𝑑 𝑉 𝑥 𝑑𝑡
−𝑚×𝑔− 1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉× 𝑉 𝑦 =𝑚× 𝑑 𝑉 𝑦 𝑑𝑡

13. Development
𝑉= 𝑉 𝑥 2 + 𝑉 𝑦 2
1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 𝑥 2 + 𝑉 𝑦 2 ×𝑉𝑥=𝑚× 𝑑 𝑉 𝑥 𝑑𝑡
−𝑚×𝑔− 1 2 ×𝜌× 𝐶 𝑑 ×𝐴× 𝑉 𝑥 2 + 𝑉 𝑦 2 × 𝑉 𝑦 =𝑚× 𝑑 𝑉 𝑦 𝑑𝑡
Because the two equations contain both the 𝑉 𝑥 and 𝑉 𝑦 variables , they are considered to be coupled. This means to further pursue the problem I needed to use a numerical solution. V Vy Vx 𝜃

14. Development Numerical solution using Euler method
We can just focus on the equation
𝑎 𝑥 t ≈ 𝑣 𝑥 𝑡+∆𝑡 − 𝑣 𝑥 (𝑡) ∆𝑡
And then multiply by ∆𝑡, resulting in
𝑎 𝑥 t ∆𝑡≈ 𝑣 𝑥 𝑡+∆𝑡 − 𝑣 𝑥 (𝑡)
Followed by adding 𝑣 𝑥 𝑡 to the other side, giving us
𝑣 𝑥 𝑡+∆𝑡 ≈ 𝑣 𝑥 𝑡 + 𝑎 𝑥 (𝑡)∆𝑡

15. Development The same can be done to find displacement
Numerical solution using Euler method
𝑣 𝑥 t = 𝑑𝑥 𝑑𝑡 ≈ ∆𝑥 ∆𝑡 = 𝑥 𝑡+∆𝑡 −𝑥(𝑡) ∆𝑡
We can just on the equation
𝑣 𝑥 t ≈ 𝑥 𝑡+∆𝑡 −𝑥(𝑡) ∆𝑡
And then multiply by ∆𝑡, resulting in
𝑣 𝑥 t ∆𝑡≈𝑥 𝑡+∆𝑡 −𝑥(𝑡)
Add 𝑥 𝑡 to the other side resulting in
𝑥 𝑡+∆𝑡 ≈𝑥 𝑡 + 𝑣 𝑥 (𝑡)∆𝑡

16. Development 𝑣 𝑦 𝑡+∆𝑡 ≈ 𝑣 𝑦 𝑡 + 𝑎 𝑦 (𝑡)∆𝑡 𝑣 𝑥 𝑡+∆𝑡 ≈ 𝑣 𝑥 𝑡 + 𝑎 𝑥 (𝑡)∆𝑡
𝑣 𝑦 𝑡+∆𝑡 ≈ 𝑣 𝑦 𝑡 + 𝑎 𝑦 (𝑡)∆𝑡
𝑦 𝑡+∆𝑡 ≈𝑦 𝑡 + 𝑣 𝑦 (𝑡)∆𝑡
𝑣 𝑥 𝑡+∆𝑡 ≈ 𝑣 𝑥 𝑡 + 𝑎 𝑥 (𝑡)∆𝑡
𝑥 𝑡+∆𝑡 ≈𝑥 𝑡 + 𝑣 𝑥 (𝑡)∆𝑡
Using these four equations we can find both the ball’s velocity, as well as the ball’s position, by assigning ∆𝑡,which signifies the change in time, which I made .1 second.
Acceleration can be found by dividing 𝑚 from
Becoming and likewise for 𝑦.
1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉×𝑉𝑥=𝑚× 𝑑𝑉𝑥 𝑑𝑡
1 2 ×𝜌× 𝐶 𝑑 ×𝐴×𝑉×𝑉𝑥 𝑚 = 𝑑𝑉𝑥 𝑑𝑡 = 𝑎 𝑥

17. Development For the air density 𝜌 I used the conventional 1.21
For the cross-sectional area of the ball I used the geometric equation 𝐴=𝜋 𝑟 2 , and where the radius of the volleyball is
The drag coefficient 𝐶 𝑑 for a volleyball is known to be .1
The mass for a volleyball is generally grams, so I used 270 grams.

18. Results

19. Results

20. Conclusion/Analysis
Upon looking at my results I perceived the range of 26⁰ (19 ⁰- 45⁰) to be relatively large. However, considering that many competitive players would not want to float the ball over the net so easily, like at 45⁰, the ideal range would be significantly smaller.
My results also made me think about spin’s importance in a serve. Though I used 7 meters per second, which equates to slightly over 15 mph, some Olympic level players serve at speeds greater than 100 mph. This means that having topspin (which accelerates the ball downward) is essential to maximizing your serve speed.

21. Areas for Further Research
Ways to improve upon this project could include…
Incorporating the impact of spin
Observe serves which are not completely straight
Analyze the impact of wind in outdoor volleyball

22. Bibliography li
_aerodynamics_of_a_new_volleyball
volleyball.html
20Trajectories.pdf