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?
what are you studying right now?
Multiple representations of functions.
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..
o sorry, it's algebra 2.
fixt.
I don't quite get what you mean. I can send a pic of the problem tho?
sure
part A and B are simple, I just don't know how to do Part C
cool. what did you get for A?
I graphed it
assuming that's right of course. lol
how...
And B is clearly 70.4
you graphed the points and ran some utility that did a quadratic fit or something?
I doubt B is 70.4
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.
So could you help me, cause I clearly don't know how to do it? lol
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
I'm so lost. what does this mean?
you have learned about parabolas right?
yes
general form for a parabola is y =ax^2 +bx +c
right?
this is a parabola...
they want you to use some or all of the points they gave you in order to find the equation of this parabola...
How do we do that?
h = at^2 +bt +c
plug them in
when t=0 h=50
50 = a*0 + b*0 +c
c=50
when t =1 h =65.1 65.1 = a*1 +b*1 +50
when t =2 h =70.4 70.4 = a*2^2 +b*2 +50 70.4 =4a + 2b +50
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)
How do I find a and b in this equation?
solve simultaneous equations: 65.1 = a +b+50 70.4 =4a + 2b +50
but how do I solve this without knowing a or b?
don't know.
probably by using a calculator that does quadratic fitting.
I just started this class. How on earth could it be this complicated?
Sorry, but I think I'll find help elsewhere
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.
oh noes.
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