SOLUTION: Hi I am confused by this question, it relates to a cork being released from a champagne bottle. The height in meters of the cork above the level ground at time t seconds is modelle

Question 204163: Hi I am confused by this question, it relates to a cork being released from a champagne bottle. The height in meters of the cork above the level ground at time t seconds is modelled into the equaiton. h= 1+8t-5t^2
h stands for height t stands for time.
I answered some eairler questions relating to this but am stuck on this next part.
Substitute h=1.65 into the equation h= 1+8t-5t^2, and rearrange this to the quadratic equation t^2-1.6t+0.13=0
I do not understand the question.
Thank you for your help.
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The height in meters of the cork above the level ground at time t seconds is modelled into the equaiton. h= 1+8t-5t^2
h stands for height t stands for time.
I answered some eairler questions relating to this but am stuck on this next part.
Substitute h=1.65 into the equation h= 1+8t-5t^2, and rearrange this to the quadratic equation t^2-1.6t+0.13=0
----------------------
1.65 = 1+8t-5t^2 gives the time that the cork is 1.65 meters above ground.
1.65 = 1+8t-5t^2
5t^2 - 8t + 0.65 = 0
You can solve for t.

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=51 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.51414284285429, 0.085857157145715. Here's your graph:

There are 2 values for t (the solver uses x) that the cork is 1.65 meters above ground. It passes 1.65 meters going up, then again going down.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think i understand what they are asking you to do.
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here's where i think:
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your equation is -5t^2 + 8t + 1 = h
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you make h = 1.65 so your equation becomes:
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-5t^2 + 8t + 1 = 1.65
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you subtract 1.65 from both sides of the equation to get:
-5t^2 + 8t + 1 - 1.65 = 0
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you multiply both sides of the equation by (-1) to get:
5t^2 - 8t - 1 + 1.65 = 0
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you combine like terms to get:
5t^2 - 8t + .65 = 0
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you divide both sides of the equation by 5 to get:
t^2 - (8/5)t + (.65/5) = 0
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this becomes:
t^2 - 1.6t + .13 = 0
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i used a small T that looks a little like a + but it's not.
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your revised equation is giving you the time it takes for the cork to reach a height of 1.65 meters.
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solving the equation of t^2 - 1.6t + .13 indicates it would take either 1.5141 seconds or .0858 seconds roughly by just truncating to the 4th decimal place.
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i would tend to believe the .0858 as being the time on the upswing.
perhaps the .1.5141 is the time it takes after it goes up and then starts coming down?
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if the trajectory is an arc that would make sense.
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i went back to the original equation and confirmed that at .0858... and 1.5141... seconds the height was exactly 1.65 meters.
that part was good.
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i checked h at .08 and at 1.52 seconds to see where it was at.
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at .08 seconds, h = 1.608 meters.
at 1.52 seconds, h = 1.608 meters again.
at 1.53 seconds, h = 1.5355 meters so it's coming down.
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apparently the cork takes .0858... seconds to reach the height of 1.65 and then hits that height again on the way down at 1.5141... seconds.
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all nice and good, but your question was already answered when i got to the point that stated the equation was equal to:
5t^2 - 8t + .65 = 0
and became:
t^2 - 1.6t + .13 = 0 after i divided both sides of the equation by 5.
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fyi,
a graph of your original equation looks like this:
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interesting that the cork starts at 1 meter before any time has elapsed, so it's only risen .65 meters when the height is 1.65 meters.
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