A particle is projected at an angle of elevation `alpha` and after t second it appears to have an elevation of `beta` as seen from the point of projection. Find the initial velocity of projection.

Explanation

To solve the problem of finding the initial velocity of a particle projected at an angle of elevation \( \alpha \), which appears to have an elevation of \( \beta \) after \( t \) seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
- The particle is projected with an initial velocity \( u \) at an angle \( \alpha \).
- After \( t \) seconds, the height \( h \) and horizontal distance \( x \) traveled by the particle need to be determined.

2. **Determine the Vertical and Horizontal Components**:
- The vertical component of the initial velocity is given by:
\[ u_y = u \sin \alpha \]
- The horizontal component of the initial velocity is given by:
\[ u_x = u \cos \alpha \]

3. **Calculate the Height \( h \)**:
- The height \( h \) of the particle after \( t \) seconds can be expressed using the kinematic equation:
\[ h = u_y t + \frac{1}{2} a_y t^2 \]
- Here, \( a_y = -g \) (acceleration due to gravity), so:
\[ h = (u \sin \alpha) t - \frac{1}{2} g t^2 \]

4. **Calculate the Horizontal Distance \( x \)**:
- The horizontal distance traveled by the particle after \( t \) seconds is:
\[ x = u_x t = (u \cos \alpha) t \]

5. **Relate the Angle of Elevation**:
- The angle of elevation \( \beta \) can be expressed using the tangent function:
\[ \tan \beta = \frac{h}{x} \]
- Substituting the expressions for \( h \) and \( x \):
\[ \tan \beta = \frac{(u \sin \alpha) t - \frac{1}{2} g t^2}{(u \cos \alpha) t} \]

6. **Simplify the Equation**:
- Rearranging gives:
\[ \tan \beta = \frac{u \sin \alpha - \frac{1}{2} g t}{u \cos \alpha} \]
- Cross-multiplying leads to:
\[ u \cos \alpha \tan \beta = u \sin \alpha - \frac{1}{2} g t \]

7. **Solve for \( u \)**:
- Rearranging the equation to isolate \( u \):
\[ u (\sin \alpha - \cos \alpha \tan \beta) = \frac{1}{2} g t \]
- Therefore, the initial velocity \( u \) can be expressed as:
\[ u = \frac{g t}{2 (\sin \alpha - \cos \alpha \tan \beta)} \]

8. **Final Expression**:
- Since \( \tan \beta = \frac{\sin \beta}{\cos \beta} \), we can substitute this into the equation:
\[ u = \frac{g t \cos \beta}{2 (\sin \alpha - \cos \alpha \frac{\sin \beta}{\cos \beta})} \]

### Final Answer:
\[ u = \frac{g t \cos \beta}{2 (\sin \alpha - \cos \alpha \tan \beta)} \]