To determine when the rock hits the water using the given equation \( h = -4.9t^2 + 10 \), we need to find when the height \( h \) is equal to 0 (the height of the water).
Setting the equation to 0 gives:
\[ 0 = -4.9t^2 + 10 \]
Solving for \( t \):
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Rearranging the equation: \[ 4.9t^2 = 10 \]
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Divide both sides by 4.9: \[ t^2 = \frac{10}{4.9} \approx 2.04 \]
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Now take the square root of both sides: \[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
To create a table, you want a reasonable scale for the range of time \( t \), which will help you see the rock's height at various times.
Since the rock hits the water at about \( t = 1.43 \) seconds, a reasonable scale for the time (t) might range from 0 to about 2 seconds, collecting values that might include increments of 0.1 seconds. This would look like:
| Time (t) | Height (h) | |----------|------------| | 0.0 | 10 | | 0.1 | 9.95 | | 0.2 | 9.80 | | 0.3 | 9.19 | | 0.4 | 8.16 | | 0.5 | 6.75 | | 0.6 | 4.96 | | 0.7 | 2.79 | | 0.8 | 0.24 | | 0.9 | -2.79 | | 1.0 | -5.76 | | 1.5 | -14.63 |
In conclusion, a reasonable scale for the range of time for the table would be 0 to 2 seconds with increments of 0.1 seconds.
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