To solve this problem, we need to calculate how long it takes for the cat to fall to the ground and how far it travels horizontally during that time.
Step 1: Calculate the time of flight
We can use the kinematic equation for vertical motion. The cat falls a vertical distance \(y = 1.1 , \text{m}\) from the table to the ground. The equation we will use is:
\[ y = \frac{1}{2} g t^2 \]
where:
- \(y\) is the vertical distance (1.1 m),
- \(g\) is the acceleration due to gravity (approximately \(10 , \text{m/s}^2\)),
- \(t\) is the time in seconds.
Rearranging the equation to solve for \(t\):
\[ 1.1 = \frac{1}{2} (10) t^2 \]
\[ 1.1 = 5 t^2 \]
\[ t^2 = \frac{1.1}{5} \]
\[ t^2 = 0.22 \]
\[ t = \sqrt{0.22} \approx 0.469 , \text{s} \]
Step 2: Calculate the horizontal distance traveled
The horizontal distance \(x\) that the cat travels while falling can be calculated using the horizontal velocity \(v_{0x} = 0.7 , \text{m/s}\) and the time of flight \(t\) we just calculated:
\[ x = v_{0x} t \]
Substituting the known values:
\[ x = (0.7 , \text{m/s}) \times (0.469 , \text{s}) \]
\[ x \approx 0.328 , \text{m} \]
Conclusion
The cat lands approximately \(0.328 , \text{m}\) away from the edge of the table.