To determine when the rock will hit the bottom of the crevasse, we need to find the time \( x \) when the function \( f(x) = -4.9x^2 + 9 \) equals \(-2\) (since the bottom of the crevasse is 2 meters below ground level). This gives us the equation:
\[ -4.9x^2 + 9 = -2 \]
Now, let's solve for \( x \):
- Add \( 2 \) to both sides:
\[ -4.9x^2 + 9 + 2 = 0 \]
which simplifies to:
\[ -4.9x^2 + 11 = 0 \]
- Rearranging gives:
\[ 4.9x^2 = 11 \]
- Now, divide both sides by \( 4.9 \):
\[ x^2 = \frac{11}{4.9} \]
- Taking the square root of both sides:
\[ x = \sqrt{\frac{11}{4.9}} \approx \sqrt{2.2449} \approx 1.4982 \]
- Rounding to the nearest tenth gives:
\[ x \approx 1.5 \]
Thus, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.