To determine when the rock will hit the bottom of the crevasse (2 meters below ground level), we need to set the equation for the height of the rock, \(y = -4.9t^2 + 9\), equal to -2 (since -2 meters is our reference point for the bottom of the crevasse).
Setting up the equation:
\[ -4.9t^2 + 9 = -2 \]
Now, we can solve for \(t\):
- Add 2 to both sides:
\[ -4.9t^2 + 9 + 2 = 0 \]
\[ -4.9t^2 + 11 = 0 \]
- Rearranging gives:
\[ -4.9t^2 = -11 \] \[ 4.9t^2 = 11 \]
- Divide both sides by 4.9:
\[ t^2 = \frac{11}{4.9} \]
- Calculating the right-hand side:
\[ t^2 \approx 2.2449 \]
- Now, take the square root of both sides:
\[ t \approx \sqrt{2.2449} \approx 1.5 \]
Thus, the rock will hit the bottom of the crevasse:
\[ \text{The rock will hit the bottom of the crevasse in } \approx 1.5 \text{ seconds.} \]
Rounding to the nearest tenth gives us:
The rock will hit the bottom of the crevasse in 1.5 seconds.