To find when the rock will hit the bottom of the crevasse, we need to solve the equation \( f(x) = -4.9x^2 + 9 \) for when \( f(x) = -2 \), since the bottom of the crevasse is 2 meters below ground level.
Set the equation equal to -2:
\[ -4.9x^2 + 9 = -2 \]
Now, let's solve for \( x \):
- Add 2 to both sides:
\[ -4.9x^2 + 9 + 2 = 0 \]
\[ -4.9x^2 + 11 = 0 \]
- Rearranging gives:
\[ 4.9x^2 = 11 \]
- Divide by 4.9:
\[ x^2 = \frac{11}{4.9} \]
- Taking the square root of both sides:
\[ x = \sqrt{\frac{11}{4.9}} \approx 1.51 \]
- Rounding to the nearest tenth gives:
\[ x \approx 1.5 \]
Thus, the rock will hit the bottom of the crevasse in approximately \( \mathbf{1.5} \) seconds.