If the max. range up an inclined plane be one-third of the max. range down the plane then the inclination of the plane to the horizontal is

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to establish the relationship between the maximum range of a projectile on an inclined plane and the angle of inclination. We are given that the maximum range up an inclined plane is one-third of the maximum range down the plane.

### Step-by-Step Solution:

1. **Understanding the Range on an Inclined Plane**:
The maximum range \( R \) of a projectile on an inclined plane can be expressed as:
\[ R_{\text{up}} = \frac{u^2}{g} \cdot \frac{1 + \sin \theta}{\cos^2 \theta} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.

2. **Maximum Range Down the Plane**:
The maximum range down the inclined plane can be expressed as:
\[ R_{\text{down}} = \frac{u^2}{g} \cdot \frac{1 - \sin \theta}{\cos^2 \theta} \]

3. **Setting Up the Relationship**:
According to the problem, the maximum range up the inclined plane is one-third of the maximum range down the plane:
\[ R_{\text{up}} = \frac{1}{3} R_{\text{down}} \]

4. **Substituting the Expressions**:
Substituting the expressions for \( R_{\text{up}} \) and \( R_{\text{down}} \):
\[ \frac{u^2}{g} \cdot \frac{1 + \sin \theta}{\cos^2 \theta} = \frac{1}{3} \left( \frac{u^2}{g} \cdot \frac{1 - \sin \theta}{\cos^2 \theta} \right) \]

5. **Simplifying the Equation**:
Cancel \( \frac{u^2}{g \cos^2 \theta} \) from both sides:
\[ 1 + \sin \theta = \frac{1}{3} (1 - \sin \theta) \]

6. **Clearing the Fraction**:
Multiply both sides by 3:
\[ 3(1 + \sin \theta) = 1 - \sin \theta \]
This simplifies to:
\[ 3 + 3\sin \theta = 1 - \sin \theta \]

7. **Rearranging the Terms**:
Bringing all terms involving \( \sin \theta \) to one side:
\[ 3 + 3\sin \theta + \sin \theta = 1 \]
\[ 4\sin \theta = 1 - 3 \]
\[ 4\sin \theta = -2 \]
\[ \sin \theta = -\frac{1}{2} \]

8. **Finding the Angle**:
The angle \( \theta \) for which \( \sin \theta = -\frac{1}{2} \) corresponds to:
\[ \theta = 210^\circ \text{ or } 330^\circ \]
However, since we are looking for the angle of inclination to the horizontal, we take:
\[ \theta = 210^\circ - 180^\circ = 30^\circ \]

### Final Answer:
The inclination of the plane to the horizontal is \( 30^\circ \).