A particle is projected up an inclined plane of inclination `beta` at an elevation `prop` to the horizontal. Find the ratio between `tan prop and tan beta`, if the particle strikes the plane horizontally.

Explanation

To solve the problem, we need to find the ratio between \(\tan \alpha\) and \(\tan \beta\) given that a particle is projected up an inclined plane at an angle \(\alpha\) to the horizontal and strikes the plane horizontally.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is projected at an angle \(\alpha\) to the horizontal.
- The inclined plane has an angle \(\beta\) with the horizontal.
- The particle strikes the inclined plane horizontally.

2. **Identify the Conditions**:
- When the particle strikes the plane horizontally, it means that at the point of impact, the vertical component of its velocity is zero.

3. **Using Kinematic Equations**:
- The time of flight for the projectile can be derived using the vertical motion equations. The time taken to reach the maximum height is given by:
\[ t = \frac{u \sin \alpha}{g} \]
- The total time of flight until it strikes the inclined plane is twice this time (up and down):
\[ T = 2t = \frac{2u \sin \alpha}{g} \]

4. **Finding the Maximum Height**:
- The maximum height \(H\) reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \alpha}{2g} \]

5. **Finding the Range on the Inclined Plane**:
- The horizontal range \(R\) of the projectile can be calculated as:
\[ R = \frac{u^2 \sin(2\alpha)}{g} \]
- However, since the particle strikes the plane horizontally, we need to consider the effective range along the inclined plane.

6. **Relating Height and Range to the Incline**:
- The height \(H\) can be related to the angle of the incline \(\beta\) using the tangent function:
\[ \tan \beta = \frac{H}{R} \]

7. **Substituting Values**:
- Substitute \(H\) and \(R\) into the equation:
\[ \tan \beta = \frac{\frac{u^2 \sin^2 \alpha}{2g}}{\frac{u^2 \sin(2\alpha)}{g}} \]
- Simplifying this gives:
\[ \tan \beta = \frac{\sin^2 \alpha}{2 \sin(2\alpha)} \]
- Since \(\sin(2\alpha) = 2 \sin \alpha \cos \alpha\), we can write:
\[ \tan \beta = \frac{\sin^2 \alpha}{4 \sin \alpha \cos \alpha} = \frac{\sin \alpha}{4 \cos \alpha} = \frac{1}{4} \tan \alpha \]

8. **Finding the Ratio**:
- Rearranging gives us:
\[ \tan \alpha = 4 \tan \beta \]
- Therefore, the ratio is:
\[ \frac{\tan \alpha}{\tan \beta} = 4 \]

### Final Answer:
The ratio between \(\tan \alpha\) and \(\tan \beta\) is:
\[ \frac{\tan \alpha}{\tan \beta} = 4 \]