# Application of quadratic functions to volleyball

## 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?

## 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!!

haruspex
Homework Helper
Gold Member
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?

lurflurf
Homework Helper
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?

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)?

haruspex
Homework Helper
Gold Member
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?

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 :(

lurflurf
Homework Helper
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)

lurflurf
Homework Helper
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.

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..,,

SteamKing
Staff Emeritus
Homework Helper

## 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?

## 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?

lurflurf
Homework Helper
^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.

haruspex
Homework Helper
Gold Member
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.

Maged Saeed
ehild
Homework Helper

## 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?