To determine when the rock hits the water using the equation \( h = -4.9t^2 + 10 \), we need to find the time \( t \) when the height \( h \) becomes 0 (the height of the water).
Setting the equation to 0:
\[ 0 = -4.9t^2 + 10 \]
Rearranging gives:
\[ 4.9t^2 = 10 \]
Dividing both sides by 4.9:
\[ t^2 = \frac{10}{4.9} \approx 2.04 \]
Taking the square root:
\[ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]
Based on this calculation, the rock will hit the water approximately 1.43 seconds after it is thrown.
When setting up a table to represent this scenario, you should have a range that covers at least from \( t = 0 \) to \( t = 2 \) seconds to capture the moment when the rock hits the water.
Therefore, a reasonable scale for the range could be:
2
This choice allows you to see the time intervals leading up to when the rock contacts the water.
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