You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. (Figure 1) Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.What minimum speed will allow the ball to clear the roof?
Express your answer in meters per second.

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?

You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. (Figure 1) Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend. What minimum speed will allow the ball to clear the roof? Express your answer in meters per second.

This problem involves determining the minimum speed required for a ball to clear a roof when thrown between two points. The solution uses kinematic equations to analyze the ball's horizontal and vertical motion. Learn how to apply velocity components and time of flight to solve this problem, suitable for high school students in grades 10-12.

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