anonymous
  • 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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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
  • anonymous
there is more to this question right?
anonymous
  • anonymous
like more words

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anonymous
  • anonymous
Make t=0. :)
anonymous
  • anonymous
The information is enough. :)
anonymous
  • anonymous
naw dude its asking for the vertex
anonymous
  • anonymous
Would I just substitute the time that we just figured out into the equation? The 2.4 seconds?
anonymous
  • anonymous
|dw:1444445933109:dw|
anonymous
  • anonymous
No you don't have to. Just make t=0. :) \(h(t) = -16(0)2 + 38.4(0) + 0.96\)
anonymous
  • anonymous
why would you do that?? is the friend .96 ft off the ground
anonymous
  • 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
  • anonymous
no not for t=0
anonymous
  • anonymous
t=o is your initial starting point not the vertex
anonymous
  • anonymous
or your y-intercept
anonymous
  • anonymous
If (0,y) <--- this is the vertex. Please stop misguiding the one who asked.
anonymous
  • anonymous
NO IT ISNT...
anonymous
  • anonymous
If you solve the equation mathway gave me, I got 0.96.
anonymous
  • anonymous
b/2a is the x coordinate and f(b/2a) is your Y
anonymous
  • anonymous
"(**HINT: The height would be the y-coordinate of the highest point.)" Which means that you need to find the y-coordinate.
anonymous
  • anonymous
Yes, it is \(\huge 0.96\). :)
anonymous
  • anonymous
MAX HIGHT WHICH IS THE Y OF YOUR VERTEX
anonymous
  • anonymous
t=0 is not a vertex
anonymous
  • anonymous
that is 0 time e=meaning the apple hasnt been thrown yet...
anonymous
  • anonymous
It is not a vertex because it is PART OF THE VERTEX. It is the x-intercept.
anonymous
  • anonymous
how can your friend catch something that hasnt been thrown?
anonymous
  • anonymous
@Nnesha Please confirm this.
anonymous
  • anonymous
no x intercept is finding the root dude...
anonymous
  • anonymous
you're really off
anonymous
  • anonymous
@pooja195
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
  • anonymous
yes
anonymous
  • anonymous
Show me how you will find the height then. You keep on complaining, but you're not showing any work.
anonymous
  • anonymous
Of course it is correct. I was the one who helped her. :P
Nnesha
  • Nnesha
`(**HINT: The height would be the y-coordinate of the highest point.)`
anonymous
  • anonymous
but now plug that number into the function
anonymous
  • anonymous
that is the hight
anonymous
  • anonymous
not t=0...
anonymous
  • anonymous
@mathway im not calling you dumb people make mistakes and this is one of yours
anonymous
  • anonymous
You would still get the same answer either way, 0.96.
anonymous
  • anonymous
yes buy he cant say thats what it is you have to explain it in the context of the problem
anonymous
  • anonymous
since its a root ofcourse it would cancle out the other 2 but in a 4th power polynomial that would fail miserably
anonymous
  • anonymous
so the fundamental way to explain the answer is V = {(b/2a),f(b/2a)} and you look for Y
anonymous
  • 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
  • anonymous
find the root
anonymous
  • anonymous
|dw:1444446732488:dw|
Nnesha
  • Nnesha
calm plz use the hint highest point = max so we need y -coordinate of the max point (max=vertex)
Nnesha
  • Nnesha
lol sorry for late reply >.< internet iz very slow -.-
anonymous
  • anonymous
lol its ok
anonymous
  • anonymous
ok hold up go back what did you get for your sec for the apple to get to him??????!!!!
anonymous
  • anonymous
the kid on the 3rd floor????
Nnesha
  • Nnesha
x coordinate of the vertex can be any number (0,y) = y-intercept :=)
anonymous
  • anonymous
\[\frac{ -38.4 \pm \sqrt{38.4^{2}-4(-16)(0.96)} }{ 2(-16) }\]
anonymous
  • anonymous
yea lol but only if it is centered on the orgin would that be a vertex
anonymous
  • anonymous
yea @tbull1221 that works
anonymous
  • anonymous
what does that give you?
anonymous
  • anonymous
\[\frac{ -38.4 \pm \sqrt{1536} }{ -32 }\]
anonymous
  • anonymous
Then you get: \[\frac{ -38.4 \pm 16\sqrt{6} }{ -32 }\]
anonymous
  • 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
  • anonymous
btw dude the other answer for the seconds it takes was wrong its 4.8sec
anonymous
  • anonymous
plz read http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html
anonymous
  • anonymous
im in 3rd year calc and 2 year stat i know how to find the vertex of a parabola...
anonymous
  • 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
  • anonymous
I highly suggest that you don't call me "dude." We are not close.
anonymous
  • anonymous
And is the seconds answer 2.4 or 4.8?
anonymous
  • anonymous
sorry im a fairly welcoming person i dont show nor mean it in any form of disrespect
anonymous
  • anonymous
the one that said "the amount of seconds for the apple to reach the friend" is 4.8 sec
Nnesha
  • 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
  • anonymous
because -b/2a =4.8
anonymous
  • anonymous
so that is your seconds or X on your vertex
anonymous
  • 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
  • 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
  • anonymous
he got offline...
anonymous
  • anonymous
do you understand why you were wrong @mathway ??
anonymous
  • anonymous
not trying to disrespect you. at this point im trying to help you
anonymous
  • anonymous
@tbull1221 cool?
anonymous
  • anonymous
go up like 5 comments and i did a step-by-step of the process
anonymous
  • anonymous
So the seconds is 4.8? And What is the answer to #5?
anonymous
  • anonymous
5 is the ft??
anonymous
  • 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
  • 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
  • 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
  • anonymous
because mathway said it was 2.4 sec but it wasnt it was 4.8
anonymous
  • anonymous
so then plug in 4.8 into the equation to get the answer to the Max height
anonymous
  • anonymous
then what you did with the quadratic equation would give you #5
anonymous
  • anonymous
Cool? @tbull1221
anonymous
  • anonymous
@tbull1221 are you still there??
anonymous
  • anonymous
aww tbull got off :(
Jhannybean
  • 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
  • anonymous
yea she left:(
Jhannybean
  • Jhannybean
time is always positive in this case, therefore \(t= \frac{6}{5} +\sqrt{\frac{3}{2}}\)

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