The Serve
Arguably the most important aspect of the volleyball game is the serve. It is the one time that the end result is almost completely in the players control; this is because if the serve is not made, the opponent receives a point. If the server is able to get the ball to travel over the net and hit the ground of the other teams court, or force the other team to “shank” the ball he earns a point for his own team. If the player serving the ball can apply physics to his serving strategy than he will be able to target weaker players on the opposing team.
Mass of Volleyball: 270g = 0.27kg
Official Court Length: 18 m
Official Court Length: 18 m
Questions:
#1. If Malik serves the ball at an angle of 30° above the horizontal at a resultant velocity of 11.056 m/s what is its initial vertical velocity? Vhi = Velocity Horizontal Initial Vvi = Velocity Vertical Initial
Solution:
Sin90/11.056 = Sin30/Vvi
Sin90(Vvi)=Sin30(11.056)
Vvi=Sin30(11.056)/Sin90
Vvi=5.5280
∴ The initial vertical velocity of the ball is 5.5280 m/s
Solution:
Sin90/11.056 = Sin30/Vvi
Sin90(Vvi)=Sin30(11.056)
Vvi=Sin30(11.056)/Sin90
Vvi=5.5280
∴ The initial vertical velocity of the ball is 5.5280 m/s
#2. What is its horizontal velocity?
Solution:
a^2 + b^2=c^2
c^2 - a^2=b^2
11.056^2 - 5.5280^2=Vhi^2
91.676=Vhi^2
√91.676=Vhi
9.5748=Vhi
#3. How long will it take the ball to hit the ground if it is served from 9.8 feet (2.987 m) high?
Solution:
Δd = -2.987m
Vi = 5.5280 m/s
Aav = -9.8m/s^2
Δt = ?
-2.987 = 5.5280Δt + 1/2(-9.8)Δt^2
0=-4.9Δt^2+5.5280Δt+2.987
-5.5280 +/- √5.5280 - 4(-4.9)(2.987)/2(-4.9)
Δt1= -0.253s Δt2=1.381s
∴The ball will hit the floor after 1.381 seconds.
#4. If Malik served the ball at an angle of 50° at a resultant velocity of 11.056m/s from 2.987m high will the ball make it over the net? (Assuming the net is 2.44m high)
Solution:
Sin90/11.056=Sin50/Vvi
Sin90(Vvi)=Sin50(11.056)
Vvi=Sin50(11.056)/Sin90
Vvi=8.469
Δd = 2.987 - 2.44m = 0.547
Vi = 8.469
Aav = -9.8m/s^2
-0.547 = 8.4694Δt + 1/2(-9.8)Δt^2
0 = -4.9Δt^2 + 8.4694Δt + 0.547
-8.4694 +/- √8.4694-4(-4.9)(0.547)/2(-4.9)
Δt1= 0.417s ✗
Δt2= 1.311s ✔
Δd = 7.11(1.311) + 1/2(0)(1.311)^2
Δd =9.3m
∴ The ball will just make it over the net because the court is 18m long, and the net is exactly in the middle of the court (9m in).