Question

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**.