A cat chases a mouse across a 0.83 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s^2. What was the cat’s speed when it slid off the table? Answer in units of m/s.

Answer 1

To find the cat's speed when it slid off the table, we can use the physics of projectile motion.

1. Determine the time it takes for the cat to fall from the table to the floor. We can use the formula for the distance fallen under gravity:

\[ h = \frac{1}{2} g t^2 \]

where:

• \( h = 0.83 , m \) (the height of the table)
• \( g = 9.81 , m/s^2 \) (the acceleration due to gravity)
• \( t \) is the time in seconds.

Rearranging the formula to solve for \( t \):

\[ t^2 = \frac{2h}{g} \]

Substituting the known values:

\[ t^2 = \frac{2 \times 0.83 , m}{9.81 , m/s^2} \] \[ t^2 = \frac{1.66}{9.81} \approx 0.169 \] \[ t \approx \sqrt{0.169} \approx 0.411 , s \]

2. Determine the horizontal speed (\( v_{x} \)) when the cat slid off the table.

The horizontal distance the cat traveled while falling to the floor is \( d = 2.4 , m \). We can relate the horizontal speed to the distance traveled and the time taken to fall:

\[ d = v_{x} \cdot t \]

Rearranging to solve for \( v_{x} \):

\[ v_{x} = \frac{d}{t} \]

Substituting in the values:

\[ v_{x} = \frac{2.4 , m}{0.411 , s} \approx 5.84 , m/s \]

Therefore, the cat’s speed when it slid off the table was approximately 5.84 m/s.