Quadratic Question: Help Solve Homework

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What do you get for h(8.8278)?In summary, the problem involves a coin being thrown from the top of a 200m building and its height being recorded at various times. Using a quadratic function, the height of the coin can be modeled over time, with a second level of differences that is constant. By setting the function equal to 0 and using the quadratic formula, we can determine when the coin will land on the ground. However, the problem involves a different planet with a different acceleration of gravity, causing some discrepancies in the calculations.
  • #1
kylepetten
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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.
 
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  • #2
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
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
zgozvrm said:
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.

zgozvrm said:
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
zgozvrm said:
First of all, you didn't show how you came up with your equation.
Mentallic said:
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.



zgozvrm said:
Secondly, it appears that the coin experiences a change in acceleration (due to air friction?)
Mentallic said:
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
zgozvrm said:
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
eumyang said:
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
kylepetten said:
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)?
 

What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it has a variable raised to the power of two. It is in the form of ax^2 + bx + c = 0, where a, b, and c are constants.

What is the quadratic formula?

The quadratic formula is a formula used to solve quadratic equations. It is x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the constants in the quadratic equation.

What are the different methods to solve a quadratic equation?

There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. The most appropriate method depends on the given equation and the preference of the solver.

How do I know if a quadratic equation has real solutions?

A quadratic equation has real solutions if the discriminant, b^2 - 4ac, is greater than or equal to 0. If the discriminant is less than 0, the equation has no real solutions and is called an imaginary or complex solution.

How can I use a quadratic equation in real life?

Quadratic equations have many real-life applications, such as calculating the trajectory of a thrown object, determining the maximum or minimum value of a function, or finding the roots of a polynomial equation in economics and finance. They are also used in engineering, physics, and other fields.

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