To solve this problem, we need to analyze the cat's motion in two directions: horizontal and vertical.

Vertical Motion:

The vertical displacement ($\Delta y$) is 3 meters below the starting point, so: $\Delta y = -3 \, \text{m}.$

We can use the kinematic equation for vertical motion: $\Delta y = v_{y0} t + \frac{1}{2} g t^2,$ where:

• $v_{y0} = 0 \, \text{m/s}$ (since the cat is not initially moving vertically),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $t$ is the time of flight.

Substituting the known values: $-3 = 0 + \frac{1}{2} (9.8) t^2,$ $t^2 = \frac{-6}{-9.8} = 0.6122,$ $t = \sqrt{0.6122} \approx 0.78 \, \text{seconds}.$

Horizontal Motion:

Now, the horizontal displacement ($\Delta x$) is 5 meters, and horizontal speed ($v_x$) is constant since there's no acceleration in the horizontal direction.

The relationship between horizontal displacement and time is: $\Delta x = v_x t.$ Solving for $v_x$: $v_x = \frac{\Delta x}{t} = \frac{5}{0.78} \approx 6.41 \, \text{m/s}.$

Rounding to the nearest whole number, the minimum horizontal speed the cat must have is: $v_x \approx 6 \, \text{m/s}.$

Answer:

The minimum horizontal speed is 6 m/s.

Would you like further details or have any questions?

Here are 5 related questions:

1. How does the height difference affect the required speed?
2. What happens if the height difference were 0 meters?
3. What is the effect of increasing the horizontal distance?
4. How does air resistance affect real-world jumps?
5. How would the time change if the gap was 10 meters instead of 5?

Tip: For projectile motion, vertical and horizontal components of motion are independent of each other.