The function [tex]\( h(t) = -4.9t^2 + h_0 \)[/tex] gives the height [tex]\( h \)[/tex] (in meters) of an object [tex]\( t \)[/tex] seconds after it falls from an initial height [tex]\( h_0 \)[/tex].

The table shows data for a pebble that fell from a cliff.

[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time } t \, (\text{s}) & \text{Height } h \, (\text{m}) \\
\hline
1 & 55.1 \\
\hline
2 & 40.4 \\
\hline
3 & 15.9 \\
\hline
\end{array}
\][/tex]

Choose the quadratic equation that models the situation.

[tex]\[ \boxed{} \][/tex]



Answer :

To choose the quadratic equation that models the situation, we need to start with the given function that describes the height [tex]\( h(t) \)[/tex] at any time [tex]\( t \)[/tex]:

[tex]\[ h(t) = -4.9t^2 + h_0 \][/tex]

Our goal is to determine the initial height [tex]\( h_0 \)[/tex] using the data provided in the table. The data points given are:

- At [tex]\( t = 1 \)[/tex] second, [tex]\( h = 55.1 \)[/tex] meters
- At [tex]\( t = 2 \)[/tex] seconds, [tex]\( h = 40.4 \)[/tex] meters
- At [tex]\( t = 3 \)[/tex] seconds, [tex]\( h = 15.9 \)[/tex] meters

Let's use these data points to find [tex]\( h_0 \)[/tex].

### Step-by-Step Solution

1. Substitute the first data point into the equation:
[tex]\[ h(1) = -4.9(1)^2 + h_0 = 55.1 \][/tex]
[tex]\[ -4.9 + h_0 = 55.1 \][/tex]
[tex]\[ h_0 = 55.1 + 4.9 \][/tex]
[tex]\[ h_0 = 60.0 \][/tex]

2. Substitute the second data point into the equation:
[tex]\[ h(2) = -4.9(2)^2 + h_0 = 40.4 \][/tex]
[tex]\[ -4.9 \cdot 4 + h_0 = 40.4 \][/tex]
[tex]\[ -19.6 + h_0 = 40.4 \][/tex]
[tex]\[ h_0 = 40.4 + 19.6 \][/tex]
[tex]\[ h_0 = 60.0 \][/tex]

3. Substitute the third data point into the equation:
[tex]\[ h(3) = -4.9(3)^2 + h_0 = 15.9 \][/tex]
[tex]\[ -4.9 \cdot 9 + h_0 = 15.9 \][/tex]
[tex]\[ -44.1 + h_0 = 15.9 \][/tex]
[tex]\[ h_0 = 15.9 + 44.1 \][/tex]
[tex]\[ h_0 = 60.0 \][/tex]

### Conclusion
From all three data points, we consistently find that the initial height [tex]\( h_0 \)[/tex] is [tex]\( 60.0 \)[/tex] meters. Therefore, the quadratic equation that models the situation is:

[tex]\[ h(t) = -4.9t^2 + 60.0 \][/tex]