This problem can be analyzed using kinematic equations. Here's a breakdown:

Step 1: Coordinate System Setup

Let’s establish the following:

• The total distance between you and your friend is $12.0 \, \text{m}$ (since each of you is $6.0 \, \text{m}$ from opposite walls).
• The roof's edge is halfway between you and your friend, which means the ball must travel at least $6.0 \, \text{m}$ horizontally before clearing the roof.
• The height of the roof is unknown, but we'll focus on ensuring the ball clears it by finding the minimum speed.

Step 2: Time of Flight for Horizontal Motion

The horizontal velocity $v_x$ is constant because there are no horizontal forces acting on the ball. The total horizontal distance traveled by the ball is 12 meters.

If $v$ is the launch speed, and $\theta$ is the angle of launch, the horizontal velocity component is: $v_x = v \cos \theta$ The time of flight $t$ can be expressed as: $t = \frac{12.0}{v \cos \theta}$

Step 3: Vertical Motion to Clear the Roof

For vertical motion, the vertical velocity component is: $v_y = v \sin \theta$ Since the throw and catch both occur at $1.0 \, \text{m}$ above the ground, the ball starts at a height of 1 meter. To clear the roof, the ball must reach a certain height at $x = 6.0 \, \text{m}$, which corresponds to half the total horizontal distance.

The position of the ball in the vertical direction is governed by the kinematic equation: $y(t) = y_0 + v_y t - \frac{1}{2} g t^2$ where $y_0 = 1.0 \, \text{m}$ is the initial height, and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

To find the minimum speed, we need to ensure the ball clears the roof at the midpoint of its horizontal trajectory, but more details are needed to calculate this. Would you like me to continue from this step, or clarify further aspects of the problem first?