- anonymous

4) What is the height of the apple when your friend catches it?
(**HINT: The height would be the y-coordinate of the highest point.)
Double click on the space below to show your work.
h(t) = -16t^2 + 38.4t + 0.96
h= height t= time (seconds)
I will fan and medal for work plus answers!

- chestercat

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- anonymous

How do I find the height? How do I switch the equation to find the height instead of the time? I can solve it just not sure how to set it up.

- anonymous

there is more to this question right?

- anonymous

like more words

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## More answers

- anonymous

Make t=0. :)

- anonymous

The information is enough. :)

- anonymous

naw dude its asking for the vertex

- anonymous

Would I just substitute the time that we just figured out into the equation? The 2.4 seconds?

- anonymous

|dw:1444445933109:dw|

- anonymous

No you don't have to. Just make t=0. :)
\(h(t) = -16(0)2 + 38.4(0) + 0.96\)

- anonymous

why would you do that?? is the friend .96 ft off the ground

- anonymous

The height we're finding is the same thing as the vertex. =_= Notice that the vertex in this case is the maximum because \(a\) is a negative.

- anonymous

no not for t=0

- anonymous

t=o is your initial starting point not the vertex

- anonymous

or your y-intercept

- anonymous

If (0,y) <--- this is the vertex. Please stop misguiding the one who asked.

- anonymous

NO IT ISNT...

- anonymous

If you solve the equation mathway gave me, I got 0.96.

- anonymous

b/2a is the x coordinate and f(b/2a) is your Y

- anonymous

"(**HINT: The height would be the y-coordinate of the highest point.)"
Which means that you need to find the y-coordinate.

- anonymous

Yes, it is \(\huge 0.96\). :)

- anonymous

MAX HIGHT WHICH IS THE Y OF YOUR VERTEX

- anonymous

t=0 is not a vertex

- anonymous

that is 0 time e=meaning the apple hasnt been thrown yet...

- anonymous

It is not a vertex because it is PART OF THE VERTEX. It is the x-intercept.

- anonymous

how can your friend catch something that hasnt been thrown?

- anonymous

@Nnesha Please confirm this.

- anonymous

no x intercept is finding the root dude...

- anonymous

you're really off

- anonymous

- anonymous

The question I posted before this was asking if a person threw an apple how many seconds would it take to reach the other person 3 stories up? We solved and got 2.4 seconds.

- anonymous

yes

- anonymous

Show me how you will find the height then. You keep on complaining, but you're not showing any work.

- anonymous

Of course it is correct. I was the one who helped her. :P

- Nnesha

`(**HINT: The height would be the y-coordinate of the highest point.)`

- anonymous

but now plug that number into the function

- anonymous

that is the hight

- anonymous

not t=0...

- anonymous

@mathway im not calling you dumb people make mistakes and this is one of yours

- anonymous

You would still get the same answer either way, 0.96.

- anonymous

yes buy he cant say thats what it is you have to explain it in the context of the problem

- anonymous

since its a root ofcourse it would cancle out the other 2 but in a 4th power polynomial that would fail miserably

- anonymous

so the fundamental way to explain the answer is V = {(b/2a),f(b/2a)} and you look for Y

- anonymous

Okay guys last question!
Using the same equation - h(t) = -16t^2 + 38.4t + 0.96
With h = height and t = time in seconds
5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground?
(**HINT: When the apple hits the ground is when the graph hits the x-axis.)

- anonymous

find the root

- anonymous

|dw:1444446732488:dw|

- Nnesha

calm plz
use the hint
highest point = max
so we need y -coordinate of the max point (max=vertex)

- Nnesha

lol sorry for late reply >.< internet iz very slow -.-

- anonymous

lol its ok

- anonymous

ok hold up go back what did you get for your sec for the apple to get to him??????!!!!

- anonymous

the kid on the 3rd floor????

- Nnesha

x coordinate of the vertex can be any number
(0,y) = y-intercept :=)

- anonymous

\[\frac{ -38.4 \pm \sqrt{38.4^{2}-4(-16)(0.96)} }{ 2(-16) }\]

- anonymous

yea lol but only if it is centered on the orgin would that be a vertex

- anonymous

yea @tbull1221 that works

- anonymous

what does that give you?

- anonymous

\[\frac{ -38.4 \pm \sqrt{1536} }{ -32 }\]

- anonymous

Then you get:
\[\frac{ -38.4 \pm 16\sqrt{6} }{ -32 }\]

- anonymous

"x coordinate of the vertex can be any number
(0,y) = y-intercept :=)"
@Nnesha
Thank you for backing up! At least I know that I'm not dumb and I don't necessarily have to say irrelevant things when I made a mistake. :)

- anonymous

btw dude the other answer for the seconds it takes was wrong its 4.8sec

- anonymous

plz read http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html

- anonymous

im in 3rd year calc and 2 year stat i know how to find the vertex of a parabola...

- anonymous

Can someone please just give me an answer for #5? My computer is about to die. I will fan and medal if i haven't already.

- anonymous

I highly suggest that you don't call me "dude." We are not close.

- anonymous

And is the seconds answer 2.4 or 4.8?

- anonymous

sorry im a fairly welcoming person i dont show nor mean it in any form of disrespect

- anonymous

the one that said "the amount of seconds for the apple to reach the friend" is 4.8 sec

- Nnesha

people make mistakes i do all the time everyday more than 100 times
it's okay we can correct it nicely :=) so please :=)

- anonymous

because -b/2a =4.8

- anonymous

so that is your seconds or X on your vertex

- anonymous

then plug that into your equation and you will get the amount of feet it took for the friend to get the apple

- anonymous

then for this what you did was fine you just plug the numbers into the quadratic equation and you will get the time it took to hit the ground

- anonymous

he got offline...

- anonymous

do you understand why you were wrong @mathway ??

- anonymous

not trying to disrespect you. at this point im trying to help you

- anonymous

@tbull1221 cool?

- anonymous

go up like 5 comments and i did a step-by-step of the process

- anonymous

So the seconds is 4.8? And What is the answer to #5?

- anonymous

5 is the ft??

- anonymous

5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground?
(**HINT: When the apple hits the ground is when the graph hits the x-axis.)

- anonymous

but both of the ones before that were wrong because the sec were wrong so the FT to make to him wouldnt be .96

- anonymous

so with the answer to -b/2a plug that number into the equation given and you will get the distance of the throw to the vertex (the friend)

- anonymous

because mathway said it was 2.4 sec but it wasnt it was 4.8

- anonymous

so then plug in 4.8 into the equation to get the answer to the Max height

- anonymous

then what you did with the quadratic equation would give you #5

- anonymous

Cool? @tbull1221

- anonymous

@tbull1221 are you still there??

- anonymous

aww tbull got off :(

- Jhannybean

\[\begin{align} h(t) = -16t^2 + 38.4t + 0.96 &\implies h(t) = -16t^2+\frac{192}{5}t+\frac{24}{25} \\&\implies h(t) =t^2 -\frac{12}{5}t -\frac{3}{50} =0 \\&\implies h(t) = \left(t^2-\frac{12}{5}t \right) =\frac{3}{50} \\&\implies h(t) = \left(t^2 -\frac{12}{5}t +\frac{36}{25} \right)=\frac{3}{50}+\frac{36}{25} \\ &\implies h(t) = \left(t-\frac{6}{5}\right)^2=\frac{3}{2} \\&\implies h(t) : ~t=\frac{6}{5}\pm\sqrt{\frac{3}{2}} \end{align}\]

- anonymous

yea she left:(

- Jhannybean

time is always positive in this case, therefore \(t= \frac{6}{5} +\sqrt{\frac{3}{2}}\)

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