Final answer:
The squirrel's velocity before hitting the ground, ignoring air resistance, is approximately 7.67 m/s. Without the airman's data, direct comparison of deceleration isn't possible, but the squirrel's deceleration upon stopping in 2.0 cm is about -294.7925 m/s².
Step-by-step explanation:
To determine a squirrel's velocity just before hitting the ground, one can apply the basic equations of motion under the force of gravity. Ignoring air resistance, we can use the formula √v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²), and h is the height (3.0 m).
For (a), the velocity (v) at the point just before impact would be calculated as follows:
v = √(2 × 9.8 m/s² × 3.0 m) = √(58.8 m²/s²) ≈ 7.67 m/s.
For (b), to compare the squirrel's deceleration with that of an airman, we need to know the airman's stopping distance and initial velocity. Deceleration (a) can be calculated using the formula a = (v² - u²) / (2 × s), where v is the final velocity (0 m/s, as it comes to a stop), u is the initial velocity, and s is the stopping distance. Without the airman's data, we cannot directly compare the deceleration but we can calculate that of the squirrel:
a = (0 - (7.67 m/s)²) / (2 × -0.02 m) = -294.7925 m/s².
The negative sign indicates that this is deceleration, as the direction of the acceleration is opposite to the direction of motion.