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.