6.0

Step 1: Determine the distance from the center of the rod to the rings. Since the rod is 2 m long, each ring is at a distance of \frac{\ell}{2} = 1 m from the center.
Step 2: Calculate the tangential velocity (v) of each ring at the moment they leave the rod. Using \( v = \omega r \), where \( \omega = 3 \text{ rad/s} \) and \( r = 1 m \), we get \( v = 3 \times 1 = 3 \text{ m/s} \).
Step 3: Determine the time (t) it takes for the rings to fall to the ground. Using the equation for free fall \( h = \frac{1}{2}gt^2 \), where \( h = 5 m \) and \( g \approx 9.8 m/s^2 \), we rearrange to find \( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{10}{9.8}} \approx 1.01 \text{ seconds} \).
Step 4: Calculate the horizontal distance traveled by the rings before they hit the ground using the formula \( d = vt \). Thus, \( d = 3 \times 1.01 \approx 3.03 m \) for one ring. Since there are two rings, the total distance between the points on the ground where the rings fall is \( 2 \times 3.03 \approx 6.06 m \), rounding to 6.0 m.
Therefore, the distance between the fall points is approximately 6.0 m.