Application of quadratic functions to volleyball

The maximum is then the constant term b in that form.In summary, The maximum height of the ball can be determined by finding the constant term in the standard form of the given function, which can be obtained by completing the square method. This is achieved by solving the quadratic equation -16t^2 + 20t + 4 = 0 and using the standard form h = -16(t-a)^2 + b. The maximum height is then the constant term b, which can be determined by finding the vertex of the parabola.
  • #1
angeli
5
1

Homework Statement


A player hits a volleyball when it is 4 ft above the ground with an initial vertical velocity of 20 ft/s (equation would be h = -16t2 + 20t + 4). What is the maximum height of the ball?

Homework Equations


quadratic formula

The Attempt at a Solution


t = -20 ±√202 - 4(-16)(4) / 2(-16)
t = -0.175390 and 1.425390

I don't think my answer is correct, but I don't know what else to do. And I don't know how to find the maximum height of the ball using the answer above :( Any help would be greatly appreciated, thank you!
 
Physics news on Phys.org
  • #2
In the most usual formulation, there are five variables in the SUVAT equations (the equations that apply for constant acceleration). There are five equations, each involving four of the variables.
Typically, you know three values and want to find a fourth. Select the SUVAT equation which involves those four variables.
Which four variables are of interest here?
Which SUVAT equation uses those four?
 
  • #3
rewrite the equation in vertex form

h = -16t2 + 20t + 4=41/4-16(t-5/8)^2

for what t does h achieve its maximum? What is the maximum?
 
  • #4
lurflurf said:
rewrite the equation in vertex form

h = -16t2 + 20t + 4=41/4-16(t-5/8)^2

for what t does h achieve its maximum? What is the maximum?

is the maximum the vertex, which in this case is (-5/8, 41/4)?
 
  • #5
lurflurf said:
rewrite the equation in vertex form

h = -16t2 + 20t + 4=41/4-16(t-5/8)^2

for what t does h achieve its maximum? What is the maximum?
Wouldn't it be simpler to use a more appropriate SUVAT equation instead?
 
  • #6
haruspex said:
Wouldn't it be simpler to use a more appropriate SUVAT equation instead?
I tried using a SUVAT equation :( s = ut + (1/2)(a)(t^2) but ahh it confused me :(
 
  • #7
angeli said:
is the maximum the vertex, which in this case is (-5/8, 41/4)?
yes, watch the minus sign, the vertex is (5/8,41/4)
 
  • #8
haruspex said:
SUVAT
I just found out what that is. Very confusing. I am not sure who SUVAT is for, it confusing, hard to remember and limited in use.
We need to try to remember 5 useless equations to see which one does not contain time
$$h_{\mathrm{final}}=h_{\mathrm{initial}}+\dfrac{v_{\mathrm{final}}^2-v_{\mathrm{initial}}^2}{2a}$$
or
$$S=S_{\mathrm{initial}}+\dfrac{V^2-U^2}{2A}$$
is the winner! Then we can figure out what each variable should be.
 
  • #9
lurflurf said:
I just found out what that is. Very confusing. I am not sure who SUVAT is for, it confusing, hard to remember and limited in use.
We need to try to remember 5 useless equations to see which one does not contain time
hfinal=hinitial+v2final−v2initial2a​
h_{\mathrm{final}}=h_{\mathrm{initial}}+\dfrac{v_{\mathrm{final}}^2-v_{\mathrm{initial}}^2}{2a}
or
S=Sinitial+V2−U22A​
S=S_{\mathrm{initial}}+\dfrac{V^2-U^2}{2A}
is the winner! Then we can figure out what each variable should be.

That is the same SUVAT equation of :

$${v_{final}}^2={v_{initial}}^2-2g\triangle y$$

I think they derive those equations just to make the motion physics easier to calculate..,,
 
  • #10
angeli said:

Homework Statement


A player hits a volleyball when it is 4 ft above the ground with an initial vertical velocity of 20 ft/s (equation would be h = -16t2 + 20t + 4). What is the maximum height of the ball?

Homework Equations


quadratic formula

The Attempt at a Solution


t = -20 ±√202 - 4(-16)(4) / 2(-16)
t = -0.175390 and 1.425390

I don't think my answer is correct, but I don't know what else to do. And I don't know how to find the maximum height of the ball using the answer above :( Any help would be greatly appreciated, thank you!

If you have an equation which relates the height of the ball as a function of time, and you want to determine the time at which the height is a maximum, you don't solve the quadratic equation. Instead, you want to determine the time tmax when dh/dt = 0.

With h = -16t2 + 20t + 4, what is dh/dt?
 
  • #11
^I don't know the starters level, but this is a common type of problem for twelve year old students who do not know about SUVAT of derivatives. Neither are need to solve the problem.
 
  • #12
lurflurf said:
I don't know the starters level, but this is a common type of problem for twelve year old students who do not know about SUVAT of derivatives. Neither are need to solve the problem.
Solving the problem requires the use of SUVAT equations, whether you know them by that name or not.
It helps if you can remember the set of 5, but it's not essential. The one that does not involve time is equivalent to conservation of mechanical energy.
In the present problem, you can use s = ut + (1/2)at2) but you need to determine t first. This can be done directly from the definition of acceleration, a = △v/△t; this is the same as another SUVAT equation.
 
  • Like
Likes Maged Saeed
  • #13
angeli said:

Homework Statement


A player hits a volleyball when it is 4 ft above the ground with an initial vertical velocity of 20 ft/s (equation would be h = -16t2 + 20t + 4). What is the maximum height of the ball?

Homework Equations


quadratic formula

The Attempt at a Solution


t = -20 ±√202 - 4(-16)(4) / 2(-16)
t = -0.175390 and 1.425390

I don't think my answer is correct, but I don't know what else to do. And I don't know how to find the maximum height of the ball using the answer above :( Any help would be greatly appreciated, thank you!
The equation h = -16t2 + 20t + 4 gives the height as function of the time. If you solve the quadratic equation -16t2 + 20t + 4=0 you get the time when the ball reaches the ground. It is the time after the ball was kicked up, so the negative time has no meaning.
But you do not need the time, you need the maximum height. If you plot the function, you get a parabola. It has the standard form h= -16(t-a)2+b, and the maximum height is b.
Find that form of the function by the method "completing the square" . You certainly have learned it.
 

FAQ: Application of quadratic functions to volleyball

What is a quadratic function?

A quadratic function is a type of mathematical equation that can be written in the form of y = ax^2 + bx + c, where a, b, and c are constants. It is a polynomial function of degree 2 and has a parabolic shape when graphed.

How are quadratic functions used in volleyball?

Quadratic functions are used in volleyball to model the path of the ball during a serve or spike. The height of the ball can be represented by the y value, and the distance the ball travels can be represented by the x value. By using a quadratic function, we can predict the trajectory of the ball and determine where it will land on the court.

What is the vertex of a quadratic function?

The vertex of a quadratic function is the highest or lowest point on the parabola. In the context of volleyball, the vertex represents the highest point the ball will reach before it begins to descend. This point is important for players to anticipate and position themselves to receive the ball.

How can quadratic functions be used to improve volleyball skills?

By understanding and applying quadratic functions, players can improve their serving and spiking skills. By predicting the trajectory of the ball, players can adjust their technique and aim to increase the chances of a successful serve or spike. Additionally, understanding the vertex of the parabola can help players anticipate where the ball will land and position themselves accordingly.

What are some limitations of using quadratic functions in volleyball?

While quadratic functions can be useful for predicting the path of the ball, they do not take into account other factors such as air resistance and spin on the ball. Additionally, the physical abilities of the players and their technique can also impact the trajectory of the ball. Therefore, quadratic functions should be used as a tool to assist in predicting the ball's path, but not relied upon as the sole determinant of its trajectory.

Similar threads

Replies
4
Views
3K
Replies
18
Views
2K
Replies
14
Views
2K
Replies
7
Views
3K
Replies
5
Views
2K
Replies
19
Views
3K
Replies
11
Views
13K
Back
Top