To calculate the maximum height reached by the rock above the roof, we need to focus on the vertical component of the motion.

Step-by-step solution:

1. Initial vertical velocity ($v_{y0}$): The rock is thrown at an angle of $\theta = 27.0^\circ$ with an initial velocity of $v_0 = 30.0 \, \text{m/s}$.

   The vertical component of the initial velocity is given by: $v_{y0} = v_0 \sin(\theta)$ Substituting the values: $v_{y0} = 30.0 \, \text{m/s} \times \sin(27.0^\circ)$

2. Acceleration ($a$): The only acceleration acting vertically is due to gravity, $a = -9.81 \, \text{m/s}^2$.

3. Maximum height condition: At the maximum height, the vertical velocity is zero, $v_y = 0$.

   Using the following kinematic equation to find the maximum height above the roof: $v_y^2 = v_{y0}^2 + 2a \Delta y$ At the maximum height, $v_y = 0$, so the equation becomes: $0 = v_{y0}^2 + 2(-9.81) \Delta y$ Solving for $\Delta y$ (the height above the roof): $\Delta y = \frac{v_{y0}^2}{2 \times 9.81}$

Now, let's calculate the vertical component and the maximum height above the roof.The maximum height reached by the rock above the roof is approximately $9.45 \, \text{m}$.

Would you like further details or have any questions about the calculations?

Here are five related questions to explore:

1. What is the total time the rock stays in the air before hitting the ground?
2. What is the total horizontal distance traveled by the rock?
3. How does the angle of projection affect the maximum height?
4. What would the maximum height be if the angle were increased to 45 degrees?
5. How does air resistance impact the motion in a real-world scenario?

Tip: Breaking down the components of velocity into horizontal and vertical parts is crucial for solving projectile motion problems efficiently.