To determine when the rock hits the ground, we need to find the time interval when the height of the rock, [tex]\( h(t) \)[/tex], changes from being positive to zero or negative. This indicates the moment when the rock reaches the ground or goes below it.
Given the provided data:
[tex]\[
\begin{array}{|c|c|}
\hline
t & h(t) \\
\hline
0 & 20 \\
\hline
0.5 & 18.8 \\
\hline
1 & 15.1 \\
\hline
1.5 & 9 \\
\hline
2 & 0.4 \\
\hline
2.5 & -10.6 \\
\hline
3 & -24.1 \\
\hline
\end{array}
\][/tex]
We observe the heights at each given time [tex]\( t \)[/tex]:
- At [tex]\( t = 0 \)[/tex] seconds, [tex]\( h(0) = 20 \)[/tex] (rock is above the ground).
- At [tex]\( t = 0.5 \)[/tex] seconds, [tex]\( h(0.5) = 18.8 \)[/tex] (rock is above the ground).
- At [tex]\( t = 1 \)[/tex] second, [tex]\( h(1) = 15.1 \)[/tex] (rock is above the ground).
- At [tex]\( t = 1.5 \)[/tex] seconds, [tex]\( h(1.5) = 9 \)[/tex] (rock is above the ground).
- At [tex]\( t = 2 \)[/tex] seconds, [tex]\( h(2) = 0.4 \)[/tex] (rock is still above the ground).
- At [tex]\( t = 2.5 \)[/tex] seconds, [tex]\( h(2.5) = -10.6 \)[/tex] (rock is below the ground).
- At [tex]\( t = 3 \)[/tex] seconds, [tex]\( h(3) = -24.1 \)[/tex] (rock is further below the ground).
Between [tex]\( t = 2 \)[/tex] seconds and [tex]\( t = 2.5 \)[/tex] seconds, the height changes from a positive value (0.4) to a negative value (-10.6). This indicates that the rock hits the ground somewhere within this interval.
Therefore, the rock hits the ground between [tex]\( 2 \)[/tex] seconds and [tex]\( 2.5 \)[/tex] seconds after it is dropped.
In conclusion:
The rock hits the ground between [tex]\(\boxed{2}\)[/tex] seconds and [tex]\(\boxed{2.5}\)[/tex] seconds after it is dropped.
