Two bodies are projected from the same point with equal velocities in such a directions that they strike on the same point on a plane whose inclination is `beta`. If `alpha` is the angle of projection of the first , the ratio of there times of flight is

Options

A. `sin ( alpha - beta) : cos alpha`
B. `cos alpha : sin ( alpha - beta)`
C. `sin (alpha + beta) : cos alpha`
D. `cos alpha : sin ( alpha + beta)`

Answer

To solve the problem, we need to find the ratio of the times of flight of two bodies projected from the same point with equal velocities, striking the same point on an inclined plane.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two bodies projected from the same point with equal initial velocities \( u \).
- The first body is projected at an angle \( \alpha \) with respect to the horizontal, and the second body is projected at an angle \( \alpha' \).
- Both bodies strike the same point on an inclined plane that is inclined at an angle \( \beta \).

2. **Using the Range Formula**:
- The range \( R \) of a projectile on an inclined plane can be expressed as:
\[ R = \frac{u^2}{g \cos^2 \beta} \left( \sin(2\alpha - \beta) - \sin(\beta) \right) \]
- For both projectiles to reach the same point, their ranges must be equal:
\[ \frac{u^2}{g \cos^2 \beta} \left( \sin(2\alpha - \beta) - \sin(\beta) \right) = \frac{u^2}{g \cos^2 \beta} \left( \sin(2\alpha' - \beta) - \sin(\beta) \right) \]

3. **Cancelling Common Terms**:
- Since \( \frac{u^2}{g \cos^2 \beta} \) is common on both sides, we can cancel it out:
\[ \sin(2\alpha - \beta) - \sin(\beta) = \sin(2\alpha' - \beta) - \sin(\beta) \]
- This simplifies to:
\[ \sin(2\alpha - \beta) = \sin(2\alpha' - \beta) \]

4. **Using the Sine Function Property**:
- From the property of the sine function, we know:
\[ \sin A = \sin B \implies A = B \text{ or } A = \pi - B \]
- Thus, we have two cases:
- Case 1: \( 2\alpha - \beta = 2\alpha' - \beta \) (which leads to \( \alpha = \alpha' \), not useful here)
- Case 2: \( 2\alpha - \beta = \pi - (2\alpha' - \beta) \)

5. **Solving for \( \alpha' \)**:
- Rearranging Case 2 gives:
\[ 2\alpha + 2\alpha' = \pi + 2\beta \]
\[ 2\alpha' = \pi - 2\alpha - 2\beta \]
\[ \alpha' = \frac{\pi}{2} - \alpha - \beta \]

6. **Finding Time of Flight**:
- The time of flight \( T \) for the first body is given by:
\[ T = \frac{2u \sin(\alpha - \beta)}{g \cos(\beta)} \]
- The time of flight \( T' \) for the second body is:
\[ T' = \frac{2u \sin(\alpha' - \beta)}{g \cos(\beta)} \]

7. **Finding the Ratio of Times of Flight**:
- The ratio \( \frac{T}{T'} \) becomes:
\[ \frac{T}{T'} = \frac{\sin(\alpha - \beta)}{\sin(\alpha' - \beta)} \]
- Substituting \( \alpha' = \frac{\pi}{2} - \alpha - \beta \):
\[ \frac{T}{T'} = \frac{\sin(\alpha - \beta)}{\sin\left(\frac{\pi}{2} - \alpha - \beta - \beta\right)} = \frac{\sin(\alpha - \beta)}{\cos(\alpha)} \]

8. **Final Result**:
- Therefore, the ratio of the times of flight is:
\[ \frac{T}{T'} = \frac{\sin(\alpha - \beta)}{\cos(\alpha)} \]