Final answer:
The squirrel's final velocity just before hitting the ground is approximately 7.7 m/s. Its deceleration over a 2.0 cm stopping distance is 1481.25 m/s². The comparison of deceleration depends on the stopping distance of the airman; if the airman's stopping distance was greater, the squirrel's deceleration is greater.
Step-by-step explanation:
Calculating the Squirrel's Velocity and Deceleration
To answer part (a) of the question, we can use the formula for the final velocity of an object under constant acceleration (gravity in this case) which is v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the fall (3.0 m). Plugging in the numbers, we get v = √(2 × 9.8 m/s² × 3.0 m) = √(58.8 m²/s²) ≈ 7.7 m/s.
For part (b), we use the deceleration formula a = (v²)/(2d), where v is the final velocity before stopping, and d is the stopping distance. The squirrel stops in a distance of 0.02 m, so its deceleration is a = (7.7 m/s)² / (2 × 0.02 m) = 1481.25 m/s². In comparison to a previous problem involving an airman, we don't have the exact numbers but we can infer whether the squirrel's deceleration is greater or less by comparing distances over which they stop. If the airman stopped over a greater distance than 0.02 m, then the squirrel's deceleration is indeed greater. Without figures for the airman’s deceleration, we can't make a detailed comparison.