- The constant second difference of \(-8\) confirms that the height data follow a quadratic relationship.
- By solving for the coefficients, the model function is determined to be \(y = -16x^2 + 150\).
- Setting \(y = 0\) to find when the object hits the ground, we calculate \(x = \sqrt{9.375}\), which is approximately \(\boxed{3.06 \text{ seconds}}\).
Explanation
- Analyzing the Data for a Quadratic Relationship
The provided data represents the height \(y\) of an object at specific times \(x\). To verify if the data shows a quadratic relationship, we examine the differences between the consecutive height values for constant time increments of \(0.5\) seconds.
- At \(x = 0, 0.5, 1, 1.5, 2, 2.5\), the heights are \(y = 150, 146, 134, 114, 86, 50\).
- The first differences (\(\Delta y\)) are:
\(146 - 150 = -4\)
\(134 - 146 = -12\)
\(114 - 134 = -20\)
\(86 - 114 = -28\)
\(50 - 86 = -36\)
- The second differences (\(\Delta^2 y\)) are:
\(-12 - (-4) = -8\)
\(-20 - (-12) = -8\)
\(-28 - (-20) = -8\)
\(-36 - (-28) = -8\)
Since the second differences are constant (\(-8\)), the data follow a quadratic relationship of the form \(y = ax^2 + bx + c\).
- Determining the Model Function
To model the data with a function \(y = ax^2 + bx + c\), we use the points from the table to determine the coefficients \(a\), \(b\), and \(c\). Using the initial height \(y = 150\) at \(x = 0\), we find \(c = 150\). Given the constant second difference is \(2a(\Delta x)^2 = -8\), where \(\Delta x = 0.5\), we have \(2a(0.25) = -8\), which simplifies to \(0.5a = -8\), or \(a = -16\). Substituting these values into the model and checking with \(x = 0.5\), we have \(146 = -16(0.5)^2 + b(0.5) + 150\), which simplifies to \(146 = -4 + 0.5b + 150\), so \(146 = 146 + 0.5b\), meaning \(b = 0\). Thus, the model is \(y = -16x^2 + 150\).
- Calculating the Time in the Air
The object is in the air until it hits the ground, which corresponds to the height \(y = 0\). We solve the quadratic equation \(0 = -16x^2 + 150\):
\[16x^2 = 150\]
\[x^2 = \frac{150}{16} = 9.375\]
Taking the positive root since time must be non-negative:
\[x = \sqrt{9.375} \approx 3.06 \text{ seconds}.\]
The object remains in the air for approximately \(3.06\) seconds.
Examples
Quadratic relationships are fundamental in physics and engineering, such as modeling the trajectory of a ball thrown into the air or the braking distance of a vehicle. For instance, if a rescue drone drops a package, its descent height relative to time follows a quadratic path determined by gravity. Understanding these models allows for precise calculations of impact times and safety distances, which are crucial for real-world logistics and safety planning.