To determine the time it takes for the penny to hit the ground, we need to examine the values of [tex]\( h(t) \)[/tex], the height of the penny at various times [tex]\( t \)[/tex] seconds after it is dropped. The height values are recorded at specific time intervals as given in the table:
[tex]\[
\begin{array}{|c|c|}
\hline
t & h(t) \\
\hline
0 & 2 \\
\hline
0.1 & 1.951 \\
\hline
0.2 & 1.804 \\
\hline
0.3 & 1.559 \\
\hline
0.4 & 1.216 \\
\hline
0.5 & 0.775 \\
\hline
0.6 & 0.236 \\
\hline
0.7 & -0.401 \\
\hline
0.8 & -1.136 \\
\hline
\end{array}
\][/tex]
To find the time when the penny hits the ground, we look for the moment when [tex]\( h(t) \)[/tex] becomes zero or negative for the first time. Observing the table, we see the following:
- At [tex]\( t = 0.6 \)[/tex] seconds, [tex]\( h(t) = 0.236 \)[/tex], which is still above the ground.
- At [tex]\( t = 0.7 \)[/tex] seconds, [tex]\( h(t) = -0.401 \)[/tex], which is below the ground.
Since the penny is on the ground precisely when [tex]\( h(t) = 0 \)[/tex], and since the height changes from a positive value ([tex]\( 0.236 \)[/tex]) at [tex]\( t = 0.6 \)[/tex] seconds to a negative value ([tex]\( -0.401 \)[/tex]) at [tex]\( t = 0.7 \)[/tex] seconds, it indicates that the penny hits the ground sometime between [tex]\( t = 0.6 \)[/tex] seconds and [tex]\( t = 0.7 \)[/tex] seconds.
Given that we are to round the answer to the nearest tenth of a second, the closest time when the penny hits the ground is at [tex]\( t = 0.7 \)[/tex] seconds.
Thus, the correct answer is:
0.7 seconds.
