A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at 20.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.
the time the baseball spends in the air (in s)
the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m)


To solve this problem, we'll break it into two main parts:

1. **Determine the time the baseball spends in the air.**
2. **Calculate the horizontal distance from the roof edge to the point where the baseball lands.**

### 1. Time the Baseball Spends in the Air

Given dataTo solve this problem, we'll break it into two main parts:

1. **Determine the time the baseball spends in the air.**
2. **Calculate the horizontal distance from the roof edge to the point where the baseball lands.**

### 1. Time the Baseball Spends in the Air

Given 
- Initial speed of the baseball (\( v_0 \)) = 3.05 m/s
- Angle of projection (\( \theta \)) = 20.0° below the horizontal
- Height of the roof edge above the ground (\( h \)) = 2.90 m
- Acceleration due to gravity (\( g \)) = 9.81 m/s²

First, we need to resolve the initial velocity into its horizontal and vertical components:
- Horizontal component: \( v_{0x} = v_0 \cos(\theta) \)
- Vertical component: \( v_{0y} = v_0 \sin(\theta) \)

Since the angle is below the horizontal, \( v_{0y} \) will be negative.

\[ v_{0x} = 3.05 \cos(20^\circ) \]
\[ v_{0y} = -3.05 \sin(20^\circ) \]

Now, we use the kinematic equation to find the time in the air. The equation for vertical motion is:
\[ y = v_{0y} t + \frac{1}{2} g t^2 \]

Where:
- \( y \) is the vertical displacement (from roof to ground, so \( y = -2.90 \) m)
- \( v_{0y} \) is the initial vertical velocity
- \( g \) is the acceleration due to gravity
- \( t \) is the time in the air

Plugging the values into the equation:
\[ -2.90 = (-3.05 \sin(20^\circ)) t + \frac{1}{2} (9.81) t^2 \]

### 2. Horizontal Distance

Once we have the time \( t \), we can calculate the horizontal distance using the horizontal velocity (\( v_{0x} \)):
\[ d = v_{0x} t \]

Now, let's calculate these values.It seems there was an issue with solving the equation for time. Let's re-evaluate the setup of the equation to ensure it's properly configured. The equation for vertical motion should be:

\[
-2.90 = v_{0y} t + \frac{1}{2} g t^2
\]

Since we know:
\[
v_{0y} = -3.05 \sin(20^\circ)
\]
and 
\[
g = 9.81 \text{ m/s}^2
\]

Substituting and solving this equation again.There was still an issue with the calculations, resulting in complex numbers, which suggests an error in interpreting the physical setup or equations. Let's revisit and correct the approach.

1. **Re-derive the equation correctly:**
   - Set up the vertical motion equation to get the correct physical roots for time.
   - Ensure we correctly solve the quadratic equation for real, positive roots.

Let’s recompute this to ensure an accurate result.The calculations now yield the following results:

- **Time the baseball spends in the air:** \( t \approx 0.883 \) seconds
- **Horizontal distance from the roof edge to where the baseball lands:** \( d \approx 2.53 \) meters

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

Here are 5 related questions for further exploration:

1. How would the time and distance change if the roof was pitched at a different angle?
2. What would be the effect of air resistance on the baseball's flight?
3. How would the problem change if the baseball was thrown upward at an angle instead?
4. What is the impact of the initial height on the flight time and distance?
5. How would this scenario differ on another planet with a different gravitational acceleration?

**Tip:** Understanding the components of motion (horizontal and vertical) separately can simplify solving projectile problems.

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at 20.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.

Solve a real-world problem involving projectile motion where a baseball rolls off a roof pitched at 20° below the horizon. Learn how to calculate the time the baseball spends in the air and the horizontal distance it travels before landing, using kinematic equations and trigonometric concepts.

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