Final answer:
Using kinematic equations, the rock reaches a maximum height of 7.5 m above the ground before it falls into a 15 m deep hole, which is not one of the options provided.
Step-by-step explanation:
To determine the maximum height reached by the rock when thrown straight up and falling into a hole, we can use the following kinematic equation for vertical motion without air resistance:
vf^2 = vi^2 + 2 * a * d
where:
- vf is the final velocity (0 m/s at the max height)
- vi is the initial velocity (21 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s^2)
- d is the displacement (maximum height reached plus the depth of the hole)
Plugging in the values, we get:
0 = (21 m/s)^2 + 2*(-9.8 m/s^2)*d
Solving for d gives us:
d = (21 m/s)^2 / (2*9.8 m/s^2)
d = 22.5 m
The maximum height reached above the ground is the total displacement minus the depth of the hole:
Height above ground = d - hole depth
Height above ground = 22.5 m - 15 m
Height above ground = 7.5 m
Therefore, the correct answer is not listed among the options provided in the question. The rock reaches a height of 7.5 m above the ground before falling into the 15 m deep hole.