Final answer:
The rock would fall approximately 54.42 meters from rest in 5.5 seconds on Mercury.
Step-by-step explanation:
To solve this problem, we can use the kinematic equation:
d = (1/2)gt^2
where d is the distance, g is the acceleration due to gravity, and t is the time. Substituting the given values, we have:
d = (1/2)(3.59 m/s^2)(5.5 s)^2
d = 0.5(3.59 m/s^2)(30.25 s^2)
d = 54.42 m
Therefore, the rock would fall approximately 54.42 meters from rest in 5.5 seconds on Mercury. Answer: A) 98.72 m.