An object is launched from the origin with a velocity of 20.0 m/s at an angle of 30.0 degrees above the horizontal. The object lands on a roof that is 4.0 m high. What is the distance along the horizontal axis when the object hits the roof?

This is a projectile motion problem, where we need to find the horizontal distance covered by the object when it lands on the roof. Let's break the problem into two components—horizontal and vertical motion—and apply the relevant equations.

### Given:
- Initial velocity (\( v_0 \)) = 20.0 m/s
- Launch angle (\( \theta \)) = 30.0°
- Height of the roof (\( y \)) = 4.0 m
- Gravitational acceleration (\( g \)) = 9.8 m/s²

### Step 1: Find the time of flight
We'll first analyze the vertical motion, where the equation of motion is:

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

Here, \( v_{0y} = v_0 \sin(\theta) \) is the vertical component of the initial velocity, and \( y = 4.0 \, \text{m} \). The initial vertical component \( v_{0y} \) can be calculated as:

\[
v_{0y} = v_0 \sin(\theta) = 20.0 \times \sin(30.0^\circ) = 20.0 \times 0.5 = 10.0 \, \text{m/s}
\]

Substitute this into the vertical motion equation:

\[
4.0 = 10.0 t - \frac{1}{2} \times 9.8 \times t^2
\]

This simplifies to:

\[
4.0 = 10.0 t - 4.9 t^2
\]

Rearrange this into a quadratic equation:

\[
4.9 t^2 - 10.0 t + 4.0 = 0
\]

### Step 2: Solve the quadratic equation for time
We can solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( a = 4.9 \), \( b = -10.0 \), and \( c = 4.0 \), we have:

\[
t = \frac{-(-10.0) \pm \sqrt{(-10.0)^2 - 4 \times 4.9 \times 4.0}}{2 \times 4.9}
\]

Let's calculate this.The two possible times are:

- \( t_1 = 1.495 \, \text{seconds} \)
- \( t_2 = 0.546 \, \text{seconds} \)

Since we are interested in the time when the object is still in the air, we take the larger time, \( t_1 = 1.495 \, \text{seconds} \).

### Step 3: Calculate the horizontal distance
Now, let's calculate the horizontal distance using the horizontal motion equation:

\[
x = v_{0x} t
\]

where \( v_{0x} = v_0 \cos(\theta) \) is the horizontal component of the initial velocity. We can calculate \( v_{0x} \) as:

\[
v_{0x} = 20.0 \times \cos(30.0^\circ) = 20.0 \times 0.866 = 17.32 \, \text{m/s}
\]

Now, calculate the horizontal distance:

\[
x = 17.32 \times 1.495 = 25.89 \, \text{meters}
\]

### Final Answer:
The horizontal distance when the object hits the roof is approximately **25.9 meters**.

Would you like further details or explanations? Here are some related questions to explore:

1. How would the result change if the object was launched at a different angle?
2. What happens if the roof height was increased or decreased?
3. How would air resistance affect this problem?
4. How can we determine the total time of flight if there was no roof?
5. What is the maximum height reached by the object?

**Tip:** Always break down projectile motion problems into horizontal and vertical components to simplify the calculations.

Learn how to solve a projectile motion problem where an object is launched at 30 degrees with an initial velocity of 20.0 m/s and lands on a roof 4.0 m high. This solution includes breaking the problem into horizontal and vertical components, applying kinematic equations, and solving a quadratic equation to find the horizontal distance. Suitable for students in Grades 9-12.

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