# Parabolic expression/vertex/velocity etc.. HELP

• Dec 24th 2009, 08:37 AM
twinbabies
Parabolic expression/vertex/velocity etc.. HELP
I have been trying to figure out this problem since two days ago... I'm stuck.. and need help. I'm missing something, I suspect maybe with the fact that a parabolic expression is ax^2 + bx + c and the one below is not set up that way... Not looking for the answers, just some steps on each question.. then hopefully it will turn the light on in my head.. Thank you

Something fired at a 45 degree angle has an initial velocity of 340 ft/sec in both vertical and horizontal directions.. the equation is then
h(t) = -16t^2 + 340t

1. What is the ordered pair which constitutes the vertex of our parabolic expression h(t)

2. How high will our projectile ascend? How long does the projectile take to gain that maximum height?

3. What are the ordered pairs corresponding to the zeros of our parabolic expression h(t) (solutions to the equation -16t^2 + 340t=0)

4. How long does it take for our projectile to reach it's destination.
• Dec 24th 2009, 09:21 AM
Raoh
Quote:

Originally Posted by twinbabies
I have been trying to figure out this problem since two days ago... I'm stuck.. and need help. I'm missing something, I suspect maybe with the fact that a parabolic expression is ax^2 + bx + c and the one below is not set up that way... Not looking for the answers, just some steps on each question.. then hopefully it will turn the light on in my head.. Thank you

Something fired at a 45 degree angle has an initial velocity of 340 ft/sec in both vertical and horizontal directions.. the equation is then
h(t) = -16t^2 + 340t

1. What is the ordered pair which constitutes the vertex of our parabolic expression h(t)

2. How high will our projectile ascend? How long does the projectile take to gain that maximum height?

3. What are the ordered pairs corresponding to the zeros of our parabolic expression h(t) (solutions to the equation -16t^2 + 340t=0)

4. How long does it take for our projectile to reach it's destination.

hello(Happy)
for the first one try to complete the square of your function.
• Dec 24th 2009, 09:38 AM
Raoh
i think you have,
$\displaystyle h(t)=-16t^2 + 340t=-(4t-\frac{85}{2})^2+900+\frac{3625}{4}=-(4t-\frac{85}{2})^2+\frac{7225}{4}$
• Dec 24th 2009, 09:40 AM
twinbabies
Quote:

Originally Posted by Raoh
hello(Happy)
for the first one try to complete the square of your function.

Ok... -16t^2 + 340t +28900

-4(4t^2 - 85t - 7225)
now I think I'm supposed to state this = 0 and divide both sides by 4... is this right? I'm not sure what happens to the -4 in front.. does it drop away or ?
Thank you so much for your help.. the light is getting brighter
• Dec 24th 2009, 09:49 AM
skeeter
Quote:

Originally Posted by twinbabies
I have been trying to figure out this problem since two days ago... I'm stuck.. and need help. I'm missing something, I suspect maybe with the fact that a parabolic expression is ax^2 + bx + c and the one below is not set up that way... Not looking for the answers, just some steps on each question.. then hopefully it will turn the light on in my head.. Thank you

Something fired at a 45 degree angle has an initial velocity of 340 ft/sec in both vertical and horizontal directions... the equation is then
h(t) = -16t^2 + 340t

1. What is the ordered pair which constitutes the vertex of our parabolic expression h(t)

2. How high will our projectile ascend? How long does the projectile take to gain that maximum height?

3. What are the ordered pairs corresponding to the zeros of our parabolic expression h(t) (solutions to the equation -16t^2 + 340t=0)

4. How long does it take for our projectile to reach it's destination.

in the vertical direction ...

$\displaystyle h(t) = -16t^2 + 340t$

note that $\displaystyle a = -16$ , $\displaystyle b = 340$ , $\displaystyle c = 0$

from your study of parabolic graphs, you should know the vertex is located at $\displaystyle t = \frac{-b}{2a}$ , and that this value of $\displaystyle t$ is the time that the projectile will be at the apex of its trajectory.

in the horizontal direction ...

$\displaystyle x(t) = 340t$ ... there is no acceleration in the horizontal direction.
• Dec 24th 2009, 09:50 AM
Raoh
now you have,
$\displaystyle h(t)=\frac{7225}{4}-16(t-\frac{85}{8})^2$
so i think the vertex is the point$\displaystyle (\frac{85}{8},\frac{7225}{4})$.
• Dec 24th 2009, 10:03 AM
masters
Quote:

Originally Posted by twinbabies
Ok... -16t^2 + 340t +28900

-4(4t^2 - 85t - 7225)
now I think I'm supposed to state this = 0 and divide both sides by 4... is this right? I'm not sure what happens to the -4 in front.. does it drop away or ?
Thank you so much for your help.. the light is getting brighter

Hi twinbabies,

$\displaystyle h(t)=-16t^2+340t$

$\displaystyle h(t)=-16(t^2-\frac{85}{4}t)$

$\displaystyle h(t)=-16\left(t^2-\frac{85}{4}t+\left(\frac{85}{8}\right)^2\right)+\ frac{7225}{4}$

$\displaystyle h(t)=-16\left(t-\frac{85}{8}\right)^2+\frac{7225}{4}$

This should answer questions 1 and 2 for you.
• Dec 24th 2009, 10:35 AM
twinbabies
Thank you all for your help. I am working on these and still really am confused. These problems are more advanced than the basics we have been learning... challenging...
So far I'm getting the ordered pair for the vertex (0, 21.25) or is this the ordered pair for the zeros? if c=0, however if completing the square then the vertex (85/8, 7225/4)...
no matter what I do I keep coming up with 10.625 seconds to gain max ht of 21.25.. is this right?
• Dec 24th 2009, 10:57 AM
masters
Quote:

Originally Posted by twinbabies
Thank you all for your help. I am working on these and still really am confused. These problems are more advanced than the basics we have been learning... challenging...
So far I'm getting the ordered pair for the vertex (0, 21.25) or is this the ordered pair for the zeros? if c=0, however if completing the square then the vertex (85/8, 7225/4)...
no matter what I do I keep coming up with 10.625 seconds to gain max ht of 21.25.. is this right?

Your vertex is (t, h(t)) = (85/8, 7225/4). This means that the maximum height of 7225/4 or 1806.25 ft. is reached in 85/8 or 10.625 seconds.

The x-intercepts (zeros) of your parabola are at t=0, and t=21.25

At t=0 seconds, the projectile is at rest. At t=21.25 seconds, the projectile has reached its destination.
• Dec 24th 2009, 12:04 PM
twinbabies
Got it
Haha... so the projectile leaves the ground at 0, reaches it's maximum height of 1806.35 in 10.625 sec and had traveled 3612.5 distance, reaching ground 0 in 21.25 sec.

Thank you so much for your help... I get it....(Rofl)