A volleyball court measures 30- feet wide by 60-feet long. The net is located 30-feet from the serve line. Assume the server stands 31 feet from the net she serves the ball. The net is approximately 7. 5 feet from the ground. Write an equation that would show the path of a served ball that would clear the net and explain how you found your equation. Then document your equation.
Can you solve it in terms of an algebraic expression?
The Correct Answer and Explanation is :
To find an equation for the path of a served volleyball that clears the net, we can model the ball’s path as a projectile motion scenario. The equation of the path of the ball is a parabola since projectiles under the influence of gravity follow a parabolic trajectory. Here’s the step-by-step process and derivation:
Assumptions and Known Values:
- Initial position: The ball is served from a point 31 feet from the net. We assume the server serves the ball from a height of 8 feet (approximately shoulder height).
- Net height: The volleyball net is 7.5 feet high.
- Distance to the net: The net is located 30 feet from the serving line, so the server is 1 foot behind this point.
- Projectile motion equation: The equation of a parabola representing the ball’s flight can be written as: [
y = ax^2 + bx + c
] where:
- ( y ) is the height of the ball at any point ( x ),
- ( x ) is the horizontal distance from the server,
- ( a ), ( b ), and ( c ) are constants to be determined.
Step-by-Step Derivation:
- At the initial position (( x = 0 )): The ball is at the server’s height of 8 feet, so ( y(0) = 8 ). This gives ( c = 8 ), so the equation becomes: [
y = ax^2 + bx + 8
] - At the net (( x = 30 )): The height of the ball must be at least 7.5 feet to clear the net, so ( y(30) = 7.5 ). Substituting ( x = 30 ) and ( y = 7.5 ) into the equation: [
7.5 = a(30)^2 + b(30) + 8
]
Simplifying:
[
7.5 = 900a + 30b + 8
]
[
-0.5 = 900a + 30b \quad \text{(Equation 1)}
] - At the point where the ball hits the ground (( x = 60 )): The ball hits the ground, so ( y(60) = 0 ). Substituting ( x = 60 ) and ( y = 0 ): [
0 = a(60)^2 + b(60) + 8
]
Simplifying:
[
0 = 3600a + 60b + 8
]
[
-8 = 3600a + 60b \quad \text{(Equation 2)}
]
Solving the System of Equations:
Now, we solve the system of two equations:
- From Equation 1: ( 900a + 30b = -0.5 )
- From Equation 2: ( 3600a + 60b = -8 )
We can multiply Equation 1 by 2 to align the ( b )-coefficients:
[
1800a + 60b = -1
]
Subtracting Equation 1 from Equation 2:
[
(3600a + 60b) – (1800a + 60b) = -8 – (-1)
]
Simplifying:
[
1800a = -7
]
[
a = -\frac{7}{1800}
]
Now substitute ( a ) back into Equation 1:
[
900\left(-\frac{7}{1800}\right) + 30b = -0.5
]
[
-\frac{7}{2} + 30b = -0.5
]
[
30b = -0.5 + \frac{7}{2}
]
[
30b = -0.5 + 3.5 = 3
]
[
b = \frac{3}{30} = \frac{1}{10}
]
Final Equation:
Thus, the equation for the path of the ball is:
[
y = -\frac{7}{1800}x^2 + \frac{1}{10}x + 8
]
This equation represents the trajectory of the volleyball served from a height of 8 feet, ensuring it clears the net at 7.5 feet when it is 30 feet away from the server. The negative coefficient for ( x^2 ) indicates the ball follows a parabolic path, initially rising and then falling.