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Quadratic Question

  • Thread starter kylepetten
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  • #1
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Homework Statement



A coin it thrown from the top of a building, and its height in metres above the ground after 0, 1, 2, 3, 4, and 5 seconds is recorded as follows: {200, 203, 202, 197, 188, 175}. algebraically determine the function that describes this situation and use it to determine when the coin lands on the ground.

Homework Equations



Nil

The Attempt at a Solution



The second level of differences is constant, so it has to be a quadratic problem.

With the help of my trusty TI calculator, I determined the function to be h(t) = -2t2 + 5t + 200

So, to determine when the penny hits the ground, I let h(t) = 0.

-2t2 + 5t + 200 = 0

From here I used the quadratic formula.

t = [ -(5) +/- [tex]\sqrt{}5^2 - 4(-2)(200)[/tex] ] / [ 2(-2) ]

t = 11.3278 and 8.8278

But, when I plug this back into the function I had, I do not get h(t) = 0

I am so very confused.

Please help. Thanks.
 

Answers and Replies

  • #2
196
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you should get a negative 8.8278, leaving the only reasonable answer as 11.3278 which when i pluuged it into your equation came out at about zero (.00089432) which is close enough to be attributed to an error in the regression, since you said you got the equation from your graphing calculator
 
  • #3
754
1
First of all, you didn't show how you came up with your equation.

Secondly, it appears that the coin experiences a change in acceleration (due to air friction?)
 
  • #4
Mentallic
Homework Helper
3,798
94
First of all, you didn't show how you came up with your equation.
He didn't, he used a calculator to do the dirty work for him.

Secondly, it appears that the coin experiences a change in acceleration (due to air friction?)
Why do you say that? the last 4 values differ by 5, then 9, then 13. Their difference in change is a constant 4 which would be expected by an object undergoing uniform acceleration.
 
  • #5
754
1
First of all, you didn't show how you came up with your equation.
He didn't, he used a calculator to do the dirty work for him.
Yes, I know. But that makes it difficult for others to help him, since can't follow his logic.



Secondly, it appears that the coin experiences a change in acceleration (due to air friction?)
Why do you say that?
We are given several times and distances, so using

[tex]D_y = D_i + V_{iy} \cdot t + \frac{1}{2} \cdot a \cdot t^2[/tex]

which can be rearranged to

[tex]V_{iy} = \frac{D_y - \frac{1}{2} \cdot a \cdot t^2 - D_i}{t}[/tex]

At time t=0 sec, the height is Di = 200m, so we know the building is 200m tall.
Given acceleration of gravity of -9.8 m/s/s,

The equation becomes

[tex]V_{iy} = \frac{D_y + 4.9 \cdot t^2 - 200}{t}[/tex]

Plugging in values t = 1 sec and Dy = 203m, you get Viy = 7.9 m/s

If 7.9 m/s was the initial vertical velocity and the acceleration was a constant -9.8 m/s/s, then all of the time/distance values would calculate correctly. But, that is not the case.
At 4 seconds, the height of the coin was recorded at 188m. Using Viy = 7.9 m/s and t = 4 sec, you come up with

[tex]D_y = 200 + (7.9) \cdot 4 - \frac{1}{2} \cdot (9.8) \cdot 4^2 = 153.2 m[/tex]
If the acceleration had been constant, this would have resulted in 188m.

Try this with any of the time/distance combinations given. It doesn't work out with constant acceleration.
 
  • #6
eumyang
Homework Helper
1,347
10
We are given several times and distances, so using

[tex]D_y = D_i + V_{iy} \cdot t + \frac{1}{2} \cdot a \cdot t^2[/tex]

which can be rearranged to

[tex]V_{iy} = \frac{D_y - \frac{1}{2} \cdot a \cdot t^2 - D_i}{t}[/tex]

At time t=0 sec, the height is Di = 200m, so we know the building is 200m tall.
Given acceleration of gravity of -9.8 m/s/s,
...
Here is the problem. You are assuming that g = 9.8 m/s2 (which is a valid assumption). However, looking at the original equation:
[tex]h(t) = -2t^2 + 5t + 200[/tex]
and comparing it with the basic position-time function:
[tex]s(t) = \frac{1}{2}gt^2 + v_0 t + s_0[/tex]

We get that g = -4 m/s2 and v0 = 5 m/s. I guess this building is on a different planet. ;)
 
Last edited:
  • #7
754
1
Here is the problem. You are assuming that g = 9.8 m/s2 (which is a valid assumption). However, looking at the original equation:
[tex]h(t) = -2t^2 + 5t + 200[/tex]
and comparing it with the basic position-time function:
[tex]s(t) = \frac{1}{2}gt^2 + v_0 t + s_0[/tex]

We get that g = -4 m/s2 and v0 = 5 m/s. I guess this building is on a different planet. ;)
Absolutely correct! The problem DOES work with an acceleration of gravity of -4 m/s2.

The problem doesn't state that this takes place on another planet, so one would assume that we're talking about Earth, where g = 9.8 m/s2. (Besides, I don't know of any other planets with 200m high buildings on them). Also, since time is based on the rotation of a planet about it's axis (and it is unlikely that this lower-gravity planet rotates at the same rate as Earth), 1 second on that planet would not be the same as 1 second on Earth.
 
  • #8
754
1
t = 11.3278 and 8.8278

But, when I plug this back into the function I had, I do not get h(t) = 0

I am so very confused.

Please help. Thanks.
Now that we've established that this problem takes place on another civilized planet with buildings and money (in the form of coins), assuming an acceleration of gravity of -4 m/s2, your work looks fine, with the exception of your missing a negative sign for one of your values of t (as pointed out by armolinasf).

What do you get for h(0)?
What do you get for h(11.3278)?
 

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