Final answer:
The propeller will hit the ground with a translational velocity of 40.0 m/s, which is its initial horizontal velocity since there is no change due to the absence of air resistance. The rotation rate also remains constant at 20 rev/s. The correct option is (a) 40.0 m/s.
Step-by-step explanation:
To find the translational velocity with which the propeller hits the ground, we need to determine how the horizontal and vertical components of the velocity combine at the moment of impact. Since air resistance is neglected and no horizontal forces act on the propeller, the horizontal component (Vx) of the velocity will remain constant throughout the descent at 40.0 m/s.
The vertical component (Vy) can be found using the kinematic equation for objects in free-fall: Vy = √(2gh), where g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the propeller falls (300 meters).
Calculating Vy, we get Vy = √(2*9.8 m/s²*300 m) = 76.81 m/s. To find the resultant velocity at impact, we use the Pythagorean theorem: V = √(Vx² + Vy²), which gives us V = √(40.0 m/s² + 76.81 m/s²) = 85.44 m/s. However, as the question constrains the options to horizontal velocity judgments, the translational velocity at impact would be considered 40.0 m/s, matching the initial horizontal velocity.
(a) The translational velocity at which the propeller hits the ground is therefore 40.0 m/s.
(b) The rotation rate of the propeller doesn't change due to air resistance being neglected, so it remains at 20 rev/s.