To tackle this problem, we need to analyze the motion of the particle projected from the cabin on the inclined plane. The key here is to understand the relationship between the maximum height attained by the particle and its range, as well as how these relate to the angle of inclination, denoted as \( r \).
Understanding the Motion
First, let's break down the scenario. The cabin is moving up the inclined plane with a constant acceleration of \( g \sin(r) \). When the particle is projected from the cabin, it has an initial velocity \( v \) with respect to the cabin, directed perpendicular to the inclined plane.
Maximum Height Calculation
The maximum height \( H \) that the particle reaches relative to the inclined plane can be calculated using the formula for vertical motion:
- The vertical component of the initial velocity is \( v \sin(r) \).
- Using the equation of motion \( H = \frac{(v \sin(r))^2}{2g} \), we can find the maximum height.
Range Calculation
Next, we need to find the range \( R \) of the particle with respect to the cabin, which is moving up the incline. The range can be derived from the horizontal component of the initial velocity:
- The horizontal component is \( v \cos(r) \).
- The time of flight \( t \) until the particle returns to the inclined plane can be calculated from the vertical motion, which gives \( t = \frac{2(v \sin(r))}{g} \).
- Thus, the range \( R \) is given by \( R = v \cos(r) \cdot t = v \cos(r) \cdot \frac{2(v \sin(r))}{g} = \frac{2v^2 \sin(r) \cos(r)}{g} \).
Equating Maximum Height and Range
According to the problem, the maximum height attained by the particle is equal to the range of the particle with respect to the cabin. Therefore, we set the two equations equal:
\( \frac{(v \sin(r))^2}{2g} = \frac{2v^2 \sin(r) \cos(r)}{g} \)
Simplifying the Equation
We can simplify this equation by canceling out common terms:
- Multiplying both sides by \( 2g \) gives us \( (v \sin(r))^2 = 4v^2 \sin(r) \cos(r) \).
- Dividing both sides by \( v^2 \) (assuming \( v \neq 0 \)) results in \( \sin^2(r) = 4 \sin(r) \cos(r) \).
Finding cotangent of r
Now, we can simplify further:
- Dividing both sides by \( \sin(r) \) (assuming \( \sin(r) \neq 0 \)) gives us \( \sin(r) = 4 \cos(r) \).
- This can be rearranged to \( \cot(r) = \frac{\cos(r)}{\sin(r)} = \frac{1}{4} \).
Thus, we find that \( \cot(r) = \frac{1}{4} \). This means that the angle \( r \) can be determined using the inverse cotangent function, which provides a clear relationship between the angle of inclination and the motion of the particle.
Final Thoughts
In summary, by analyzing the motion of the particle projected from the cabin and equating the maximum height to the range, we derived that \( \cot(r) = \frac{1}{4} \). This approach not only highlights the principles of projectile motion but also emphasizes the importance of understanding the relationships between different components of motion in physics.
