To determine when the rock hits the water, we need to find the time \( t \) when the height \( h \) equals 0 (i.e., \( h = 0 \) when it hits the lake). The equation given is:
\[ h = -4.9t^2 + 10 \]
Setting \( h = 0 \):
\[ 0 = -4.9t^2 + 10 \]
Solving for \( t \):
\[ 4.9t^2 = 10 \]
\[ t^2 = \frac{10}{4.9} \]
\[ t^2 \approx 2.04 \]
\[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
To create a table that shows the height of the rock at different time intervals, we need to consider a reasonable scale for the time (the domain) and then calculate the corresponding heights.
Since the time it takes for the rock to hit the water is approximately 1.43 seconds, a reasonable scale for the range of time could be:
- A scale of 2 seconds would fit well (showing from 0 to at least 2 seconds).
- A scale of 5 seconds would be too large, as it would not provide much detail around the critical time of 1.43 seconds.
- A scale of 10 seconds would also be overly large.
- A scale of 1 second would be too granular since we only need to measure up to about 1.43 seconds.
Thus, the most reasonable option for the range (time intervals) would be:
d. 2 seconds