To find out when the rock hits the bottom of the crevasse, we need to solve the equation
\[ y = -4.9t^2 + 9 \]
for \(y = -2\) (since the bottom of the crevasse is 2 meters below ground level, which is at 0). Therefore, we set the equation equal to -2 and solve for \(t\):
\[ -4.9t^2 + 9 = -2 \]
First, we can simplify this by adding 2 to both sides:
\[ -4.9t^2 + 11 = 0 \]
Now, rearranging gives:
\[ -4.9t^2 = -11 \]
Dividing both sides by -4.9:
\[ t^2 = \frac{11}{4.9} \]
Calculating \(t^2\):
\[ t^2 \approx 2.2449 \]
Taking the square root of both sides gives:
\[ t \approx \sqrt{2.2449} \approx 1.5 \text{ seconds} \]
Therefore, rounding to the nearest tenth, the rock will hit the bottom of the crevasse in:
\[ \boxed{1.5} \text{ seconds.} \]