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The table relates to a function \(h(t)\) that models the height of a rock \(t\) seconds after it is dropped.
\begin{tabular}{|c|c|}
\hline
[tex]$t$[/tex] & [tex]$h(t)$[/tex] \\
\hline
0 & 20 \\
\hline
0.5 & 18.8 \\
\hline
1 & 15.1 \\
\hline
1.5 & 9 \\
\hline
2 & 0.4 \\
\hline
2.5 & -10.6 \\
\hline
3 & -24.1 \\
\hline
\end{tabular}

When does the rock hit the ground?

The rock hits the ground between [tex]\(\boxed{2 \text{ seconds}}\)[/tex] and [tex]\(\boxed{2.5 \text{ seconds}}\)[/tex] after it is dropped.


Sagot :

Let's solve the problem step-by-step.

1. Understand the Problem:
- The table gives the height \( h(t) \) of a rock at various times \( t \).
- We need to determine when the rock hits the ground, which happens when the height \( h(t) \) is zero or below.

2. Analyze the Data:
- The times and corresponding heights are given as follows:
- \( t = 0 \): \( h(t) = 20 \)
- \( t = 0.5 \): \( h(t) = 18.8 \)
- \( t = 1 \): \( h(t) = 15.1 \)
- \( t = 1.5 \): \( h(t) = 9 \)
- \( t = 2 \): \( h(t) = 0.4 \)
- \( t = 2.5 \): \( h(t) = -10.6 \)
- \( t = 3 \): \( h(t) = -24.1 \)

3. Determine Where the Height Falls Below Zero:
- We need to find the interval in the data where the height changes from above zero to at or below zero.

4. Compare Heights:
- \( h(2) = 0.4 \) (above zero)
- \( h(2.5) = -10.6 \) (below zero)

5. Conclusion:
- The rock hits the ground in the interval where the height changes from positive to negative, which occurs between \( t = 2 \) seconds and \( t = 2.5 \) seconds.

Answer:

The rock hits the ground between [tex]\( 2 \)[/tex] seconds and [tex]\( 2.5 \)[/tex] seconds after it is dropped.
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