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Mathematics 22 Online
OpenStudy (anonymous):

The height of a rocket at different times after it was fired was 50.0m at 0s, 65.1m at 1s, 70.4m at 2s, 65.9m at 3s, 51.5m at 4s, and 27.5m at 5 seconds. How long will the rocket stay in the air?

OpenStudy (anonymous):

what are you studying right now?

OpenStudy (anonymous):

Multiple representations of functions.

OpenStudy (anonymous):

many ways to solve this, I still don't have any idea what you're studying, so I'll just assume you're in pre-calc. or something and are learning about parabolas..

OpenStudy (anonymous):

o sorry, it's algebra 2.

OpenStudy (anonymous):

fixt.

OpenStudy (anonymous):

I don't quite get what you mean. I can send a pic of the problem tho?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

OpenStudy (anonymous):

part A and B are simple, I just don't know how to do Part C

OpenStudy (anonymous):

cool. what did you get for A?

OpenStudy (anonymous):

I graphed it

OpenStudy (anonymous):

assuming that's right of course. lol

OpenStudy (anonymous):

how...

OpenStudy (anonymous):

And B is clearly 70.4

OpenStudy (anonymous):

you graphed the points and ran some utility that did a quadratic fit or something?

OpenStudy (anonymous):

I doubt B is 70.4

OpenStudy (anonymous):

O no, I thought it was asking the best way to show the data, so my answer was to graph it. But I didn't actually do that.

OpenStudy (anonymous):

So could you help me, cause I clearly don't know how to do it? lol

OpenStudy (anonymous):

h = at^2 +bt +c at t= 0 h=50 so c=50 at t=1, h=65.1 so 65.1 = a +b +50 at t=2, h=70.4m so 70.4 = 4a +2b +50 solve for a and b and you have h(t) use it to find t when h(t) =0

OpenStudy (anonymous):

I'm so lost. what does this mean?

OpenStudy (anonymous):

you have learned about parabolas right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

general form for a parabola is y =ax^2 +bx +c

OpenStudy (anonymous):

right?

OpenStudy (anonymous):

this is a parabola...

OpenStudy (anonymous):

they want you to use some or all of the points they gave you in order to find the equation of this parabola...

OpenStudy (anonymous):

How do we do that?

OpenStudy (anonymous):

h = at^2 +bt +c

OpenStudy (anonymous):

plug them in

OpenStudy (anonymous):

when t=0 h=50

OpenStudy (anonymous):

50 = a*0 + b*0 +c

OpenStudy (anonymous):

c=50

OpenStudy (anonymous):

when t =1 h =65.1 65.1 = a*1 +b*1 +50

OpenStudy (anonymous):

when t =2 h =70.4 70.4 = a*2^2 +b*2 +50 70.4 =4a + 2b +50

OpenStudy (anonymous):

you can find a and b now, so you know what the answer to part A is... (you have an equation that models the height of the rocket as a function of time)

OpenStudy (anonymous):

How do I find a and b in this equation?

OpenStudy (anonymous):

solve simultaneous equations: 65.1 = a +b+50 70.4 =4a + 2b +50

OpenStudy (anonymous):

but how do I solve this without knowing a or b?

OpenStudy (anonymous):

don't know.

OpenStudy (anonymous):

probably by using a calculator that does quadratic fitting.

OpenStudy (anonymous):

I just started this class. How on earth could it be this complicated?

OpenStudy (anonymous):

Sorry, but I think I'll find help elsewhere

OpenStudy (anonymous):

I think they want you to actually find the equation on your own however... since the other way is not really a test of anything but whether or not you have a quadratic fitting program on your calculator.

OpenStudy (anonymous):

oh noes.

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