To determine when The Rock will hit the bottom of the crevasse, we need to find the time \( t \) when the position \( Y \) is equal to \(-2\) meters (since it's 2 meters below ground level). The equation given is:
\[ Y = -4.9t^2 + 9 \]
We set \( Y = -2 \) and solve for \( t \):
\[ -2 = -4.9t^2 + 9 \]
Now, rearranging the equation gives us:
\[ -4.9t^2 = -2 - 9 \]
\[ -4.9t^2 = -11 \]
Dividing both sides by \(-4.9\):
\[ t^2 = \frac{11}{4.9} \]
Calculating the right-hand side:
\[ t^2 \approx 2.2449 \]
Now, taking the square root of both sides to find \( t \):
\[ t \approx \sqrt{2.2449} \approx 1.5 \text{ seconds} \]
Thus, The Rock will hit the bottom of the crevasse approximately \( 1.5 \) seconds after the fall. Rounding to the nearest tenth of a second, the answer remains 1.5 seconds.