Question

A rock is launched from a cannon its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.
After 1 second, the rock is 186 feet in the air after 2 seconds, it is 344 in the air.
Find the height, in feet,of the rock after 8 seconds in the air.

Answer 1

(You are basically solving for a,b, and c and forming a new quadratic equation.

Answer 2

@mathmale @Nurali

Answer 3

something like this

-a*x^2+v*x

you have to find a and v

Answer 4

here is the data you do know

-a*1^2+v*1=186
-a*2^2+v*2=344

SOlve that system somehow.

Answer 5

I got the answer already it worked :D hehehe

Answer 6

See, I have the formula, but it takes an unnecessary amount of time to use. I was actually looking for a shorter way to solve this problem,and this may be it..

Answer 7

@timo86m "I got the answer already it worked :D hehehe" is definitely not an appropriate comment when it is YOUR responsibility to help the other person find the correct answer herself.

Answer 8

@mathmale I haven't given her the answer

Answer 9

I am just letting her know that my theory for solving it works. Now it is up to her to apply it.

Answer 10

This is basically a physics problem, but is also seen in Math as well. If it's Math, you were probably given a model for the height of an object thrown straight up in the air as a function of t. Right?

One such form of that wuold be \[h(t)=v _{0}t-\frac{ 1 }{ 2 }g*t^2\] where g is the acceleration due to gravity. You are given two pairs of data (t, h) that satisfy this equation, and you'll need to find the values of the initial velocity (v0) and of the acceleration due to gravity (g).

Answer 11

Note that this IS a quadratic formula.

I'm tempted to "cheat" a little and let g=-32 ft/(sec^2), substitute the coordinates from either point, and calculate v0 (the initial velocity). What do you think?

Answer 12

it is a quadratic

Answer 13

I was not given a model, but a quadratic equation: ax^2+bx+c. Thanks for your help, I like discovering new ways to solve this problem.

Answer 14

Maybe we start from the beginning? :)
I know all about cannons :3
Hey @alex-linne we know that h(x) is quadratic, so yeah, let's do that, the height is h, and time in seconds is x.
\[\Large h(x) = \color{red}ax^2 +\color{blue}bx + \color{green}c\]

Answer 15

@PeterPan it is easier to assume the cannon is at 0 initially os make c=0 just my assumption.

Answer 16

Alex my way is the best. :D

Answer 17

All in good time, Timo :P
There doesn't seem to be a reason to rush. So @alex-linne do we proceed?

Answer 18

Better. Okay, we can assume that at the beginning, the height of the rock (who the heck shoots rocks from cannons) is zero. After all, at time zero, like, just before it's fired, the rock is still on the ground... right? :)

Answer 19

yeess.

Answer 20

Okay. In Maths language, that amounts to
\[\Large h(0) = 0\]
OR
\[\Large h(0) = \color{red}a(0)^2 + \color{blue}b(0) + \color{green}c = \color{green}c = 0\]

Following?

Doing this gives you a reason for c being zero... instead of just assuming it.

So... do you follow? ^.^

Answer 21

not true cuzz you can be firing from a cliff. SO it still an assumption but it works.

Answer 22

Right. Nothing is happening at this point, so is that why there are zeros?

Answer 23

Nothing happening at time zero (x = 0).

So that narrows down your quadratic equation to
\[\large h(x) = \color{red}ax^2 + \color{blue}bx\]

Answer 24

simply put it is an assumption we made and it is correct. Only if they where to have told us we firing from a cliff of height h then we can plug it in.

Answer 25

ahh.

Answer 26

@TIMO86M: may i suggest that we focus on solving this problem in the shortest and most effective possible way, and only then offer alternative points of view AFTER that?

Answer 27

@alex-linne
The key to these word-problems is to know how to translate the words from English to Maths-Language :D

So, translating "Height is zero, when time is zero" gives h(0) = 0

What about "Height is 186 after 1 second" ?

Answer 28

I got 186=a+b+c..

Answer 29

Sure. But we already know that c = 0 :P

Answer 30

or at least, assumed it :)

Answer 31

So... simplify?

Answer 32

ah.. .-.

Answer 33

Just replace c with zero. Now, and from now on :D
186 = a + b + 0 = a+b

186 = a + b

We leave it at that, for now...
What about "Height is 344 after 2 seconds" ?

Answer 34

344=4a+2b? :o

Answer 35

Nicely done XD

Answer 36

Okay.
a+b = 186
4a + 2b = 344

A system of linear equations.

Can you solve it now? ^.^

Answer 37

Not really ..haha. I got a strange answer .-.

Answer 38

Really? How strange?
What did you get?

Answer 39

Wait, nevermind, I did something wrong >.>

Answer 40

Keep going... the answer's pretty nice-looking (mind you, whatever you get for a and b, those won't be the final answers yet )

Answer 41

I'm stuck at 158=3a+b.. >__>

Answer 42

You're going to do that, are you?
I suggest you multiply the first equation by 2 first, before doing anything of the sort :P

Answer 43

2a + 2b = 372

Answer 44

Ah, and then subtract that from 344=4a+2b? :o

Answer 45

Yup ^.^

Answer 46

a=14,so then what?

Answer 47

I don't think so :P

Answer 48

2a + 2b = 372
4a + 2b = 344

(4a + 2b) - (2a + 2b) = 344 - 372

Careful now :)

Answer 49

ehhh ha ha..?

Answer 50

How did you get a = 14?

Answer 51

Show steps. Then we'll see where you made the mistake.