Final answer:
The velocity of the rock just before it strikes the ground is greater than 8 m/s.option D is correct.
Step-by-step explanation:
The velocity of the rock just before it strikes the ground can be determined using the equations of motion. In this case, since the rock is thrown straight down, it experiences acceleration due to gravity in the downward direction. The initial velocity is 8 m/s and the distance traveled is 30 m. We can use the equation:
Vf^2 = Vi^2 + 2ad
Where Vf is the final velocity, Vi is the initial velocity, a is acceleration, and d is distance. Solving for Vf:
Vf^2 = (8 m/s)^2 + 2(-9.8 m/s^2)(30 m)
Vf^2 = 64 m^2/s^2 - 588 m^2/s^2
Vf^2 = -524 m^2/s^2
Since velocity is a vector quantity, the answer can be either positive or negative. However, the magnitude of the velocity is greater than 8 m/s. Therefore, the correct answer is B) Greater than 8 m/s.