Question

Need help on word problem.

Answer 1

This is the word problem and what I have is what I have on there so far
I'm stuck

Answer 2

your trying to solve for t so you need to get t on its own side of the equation by itself.

Answer 3

h = Vo*t + 0.5g*t^2
Vo = 192 Ft/s
g = -32Ft/s^2
h = 432 Ft.
Solve for t(Use Quadratic formula).

h max = -(Vo^2)/2g = -(192^2)/-64 =
576 Ft.
what math class are you in?

Answer 4

@briana.img

Answer 5

sorry i was called
but i'm taking algebra 2

Answer 6

im in algebra 2 also. i actually did this problem a couple of weeks ago and that was the answer.

Answer 7

@kourtney2014 where'd you get 0.5 from?

Answer 8

it doesn't look like the same problem??

Answer 9

ill explain

Answer 10

i'm really confused lol

Answer 11

@johnweldon1993 can you please help?

Answer 12

I was gonna let @kourtney2014 explain first :)

Answer 13

@johnweldon1993 she seems to have left??

Answer 14

Okay :)

so we have

\[\large h = 192t - 16t^2\]

we want to know t when h = 432

\[\large 432 = 192t - 16t^2\]

If you look, everything is divisible by 16...so we can bring these numbers smaller a bit
Divide everything by 16

\[\large 27 = 12t - t^2\]

Looks better to me

Now we solve for ;t;...we can use the quadratic formula for that

\[\large -t^2 + 12t - 27 = 0\]

\[\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This will solve for 't'

Answer 15

ooooh okay!! thank you!!! didn't know you could take apart 16 from the t^2

Answer 16

Hey! that got deleted for some reason?? :O

grr gotta type it out again! lol

Answer 17

no it didn't get deleted?

Answer 18

The part b) part? You still see it?

Answer 19

@johnweldon1993 ooooh no i never saw it
sorry about that
but i got 3 and 9 for the answers for the quadratic formula

Answer 20

i'll plug them in and see which one is correct?

Answer 21

wait nevermind they both work when i plug them back into the original equation lol

Answer 22

They both will be :)

What does that mean though?? :O

Answer 23

Hint* Think about the path this is traveling :)



As it is going up...at t = 3...it will hit 432 feet

and then again at 9...as it is falling back down...it will hit it again!

Answer 24

Well I know after 3 seconds it hits the heigh of 432 but what is the maximum height that the rocket reaches??

Answer 25

9 has to do with that?

Answer 26

Lol okay good part 1 done...yes t = 3

Now!!!!!! part b) forget everything we've done lol

Answer 27

But doesn't it have to do with what we've done? The maximum point on that graph you made would be the answer?

Answer 28

No jk haha...remember the equation

\[\large -t^2 + 12t - 27 = 0\]

Answer 29

We have to substitute 9?

Answer 30

And yes, the maximum point would be the answer...but what would that be?

Lets look at that little graph I did



Kinda makes sense that it would be halfway between 3 and 9 in terms of time right?

Answer 31

Yeah!!

(((sorry it takes me a while to reply
open study is so glitchy i have to refresh every time i wanna reply)))

Answer 32

I know same here lol

So that would be at time = 6 right?

What would your equation *original equation* come out to at t= 6??

Answer 33

I get 9!!

Answer 34

Work with that original equation

\[\large h=192t−16t^2\]
plug in t = here :)

Answer 35

Remember everything we have done has been from this beginning equation

To find out t at 432...we just made h = 432

Now we want to KNOW 'h' now what we know at t = 6 is the max height...so from that...just plug in 6 for 't' and solve for 'h'

Answer 36

576!!

Answer 37

BAM!!! you got it :D