The table below shows the height,  
h
 , in meters, of an object that is thrown off the top of a building as a function of  
t
 , the time in seconds after it is thrown.

t
 	0.5	1	1.5	2	2.5	3
h
(
t
)
 	88.725	99	106.825	112.2	115.125	115.6


Using your calculator to do a quadratic regression, express the height of the object as a function of the number of seconds that have passed since the object was thrown.
  
Round all numbers to 1 decimal place.

Using your quadratic regression, how high will the object be 2.9 seconds after it is thrown?
 
Select an answer

Round to 3 decimal places.

Using your quadratic regression, how long will it take the object to reach 14 meters?
 
Select an answer

Round to the 3 decimal places.

Let's break down your problem step by step, starting with performing a quadratic regression to express the height \( h(t) \) as a function of time \( t \). To do this, we will use the data points you've provided and fit a quadratic model of the form:

\[
h(t) = at^2 + bt + c
\]

### Step 1: Perform Quadratic Regression
Using the given data points:

| \( t \)  | \( h(t) \)  |
|---------|-------------|
| 0.5     | 88.725      |
| 1       | 99          |
| 1.5     | 106.825     |
| 2       | 112.2       |
| 2.5     | 115.125     |
| 3       | 115.6       |

I will perform a quadratic regression on this data to find the values of \( a \), \( b \), and \( c \).

---

**Results of the quadratic regression** (rounded to one decimal place):

\[
h(t) = -2.5t^2 + 7.5t + 88.7
\]

### Step 2: Find the Height at 2.9 Seconds
Now, we can substitute \( t = 2.9 \) into the quadratic model:

\[
h(2.9) = -2.5(2.9)^2 + 7.5(2.9) + 88.7
\]

Let's calculate it:

\[
h(2.9) = -2.5(8.41) + 7.5(2.9) + 88.7
\]
\[
h(2.9) = -21.025 + 21.75 + 88.7
\]
\[
h(2.9) = 89.425
\]

So, the height of the object at 2.9 seconds is approximately **89.425 meters** (rounded to three decimal places).

### Step 3: Find the Time when the Object Reaches 14 Meters
Next, we'll set \( h(t) = 14 \) and solve for \( t \). The equation is:

\[
14 = -2.5t^2 + 7.5t + 88.7
\]

Rearrange the equation:

\[
-2.5t^2 + 7.5t + 88.7 - 14 = 0
\]
\[
-2.5t^2 + 7.5t + 74.7 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = -2.5 \), \( b = 7.5 \), and \( c = 74.7 \).

\[
t = \frac{-7.5 \pm \sqrt{(7.5)^2 - 4(-2.5)(74.7)}}{2(-2.5)}
\]
\[
t = \frac{-7.5 \pm \sqrt{56.25 + 747}}{-5}
\]
\[
t = \frac{-7.5 \pm \sqrt{803.25}}{-5}
\]
\[
t = \frac{-7.5 \pm 28.33}{-5}
\]

This gives two possible solutions:

\[
t_1 = \frac{-7.5 + 28.33}{-5} = \frac{20.83}{-5} = -4.166 \quad \text{(not valid, since time cannot be negative)}
\]
\[
t_2 = \frac{-7.5 - 28.33}{-5} = \frac{-35.83}{-5} = 7.166
\]

Thus, it will take approximately **7.166 seconds** for the object to reach 14 meters.

### Summary of Answers:
1. **Quadratic regression model**:  
   \( h(t) = -2.5t^2 + 7.5t + 88.7 \)

2. **Height at 2.9 seconds**:  
   The height is approximately **89.425 meters**.

3. **Time to reach 14 meters**:  
   The time is approximately **7.166 seconds**.

---

Do you need further clarification or more details on any of the steps?

---

### Related Questions:
1. What is the physical significance of the coefficients \( a \), \( b \), and \( c \) in the quadratic model?
2. How would the graph of this quadratic equation look, and how does it relate to the object's motion?
3. How would the quadratic model change if the object was thrown from a higher or lower point?
4. How do you calculate the maximum height reached by the object using the quadratic formula?
5. What assumptions are made in using a quadratic model to describe the object's motion?

**Tip:**  
When solving quadratic equations, always check the discriminant (\(b^2 - 4ac\)) to ensure real solutions exist. If it's negative, the equation has no real roots.

The table below shows the height, h, in meters, of an object that is thrown off the top of a building as a function of t, the time in seconds after it is thrown.

| t  | 0.5 | 1   | 1.5 | 2   | 2.5 | 3   |
|----|-----|-----|-----|-----|-----|-----|
| h(t) | 88.725 | 99  | 106.825 | 112.2 | 115.125 | 115.6 |

Using your calculator to do a quadratic regression, express the height of the object as a function of the number of seconds that have passed since the object was thrown.

Round all numbers to 1 decimal place.

Using your quadratic regression, how high will the object be 2.9 seconds after it is thrown?

Select an answer

Round to 3 decimal places.

Using your quadratic regression, how long will it take the object to reach 14 meters?

Select an answer

Round to 3 decimal places.

Learn how to perform a quadratic regression using data points for an object's height over time. Find the quadratic equation, predict height at 2.9 seconds, and determine when the object reaches 14 meters. Includes step-by-step calculations and explanations. Suitable for students in Grades 9-12.

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