A particle is projected from a point on the level ground and its height is `h` when at horizontal distances `a and 2 a` from its point of projection. Find the velocity of projection.

Explanation

To solve the problem, we need to find the velocity of projection \( u \) of a particle that reaches a height \( h \) at horizontal distances \( a \) and \( 2a \) from the point of projection.

### Step-by-Step Solution:

1. **Understand the Projectile Motion Equation**:
The equation of the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
where \( y \) is the height, \( x \) is the horizontal distance, \( g \) is the acceleration due to gravity, \( u \) is the initial velocity, and \( \theta \) is the angle of projection.

2. **Set Up the Equations for Given Points**:
For the point at distance \( a \):
\[ h = a \tan \theta - \frac{g a^2}{2 u^2 \cos^2 \theta} \quad \text{(1)} \]
For the point at distance \( 2a \):
\[ h = 2a \tan \theta - \frac{g (2a)^2}{2 u^2 \cos^2 \theta} \quad \text{(2)} \]

3. **Rearranging the Equations**:
Rearranging equation (1):
\[ h + \frac{g a^2}{2 u^2 \cos^2 \theta} = a \tan \theta \]
Rearranging equation (2):
\[ h + \frac{2g a^2}{u^2 \cos^2 \theta} = 2a \tan \theta \]

4. **Substituting for \( \tan \theta \)**:
From equation (1), we can express \( \tan \theta \):
\[ \tan \theta = \frac{h + \frac{g a^2}{2 u^2 \cos^2 \theta}}{a} \]
From equation (2):
\[ \tan \theta = \frac{h + \frac{2g a^2}{u^2 \cos^2 \theta}}{2a} \]

5. **Equating the Two Expressions for \( \tan \theta \)**:
Set the two expressions for \( \tan \theta \) equal to each other:
\[ \frac{h + \frac{g a^2}{2 u^2 \cos^2 \theta}}{a} = \frac{h + \frac{2g a^2}{u^2 \cos^2 \theta}}{2a} \]
Cross-multiplying gives:
\[ 2(h + \frac{g a^2}{2 u^2 \cos^2 \theta}) = h + \frac{2g a^2}{u^2 \cos^2 \theta} \]

6. **Simplifying the Equation**:
This simplifies to:
\[ 2h + \frac{g a^2}{u^2 \cos^2 \theta} = h + \frac{2g a^2}{u^2 \cos^2 \theta} \]
Rearranging gives:
\[ h = \frac{g a^2}{u^2 \cos^2 \theta} \]

7. **Finding \( u^2 \)**:
Rearranging for \( u^2 \):
\[ u^2 = \frac{g a^2}{h \cos^2 \theta} \quad \text{(3)} \]

8. **Finding \( \tan \theta \)**:
From the earlier expression for \( \tan \theta \):
\[ \tan \theta = \frac{3h}{2a} \]
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ \sec^2 \theta = 1 + \left(\frac{3h}{2a}\right)^2 \]

9. **Substituting back into Equation (3)**:
Substitute \( \sec^2 \theta \) back into equation (3):
\[ u^2 = \frac{g a^2}{h} \left(1 + \frac{9h^2}{4a^2}\right) \]
Simplifying gives:
\[ u^2 = \frac{g a^2}{h} \cdot \frac{4a^2 + 9h^2}{4a^2} \]

10. **Final Expression for \( u \)**:
Taking the square root gives:
\[ u = \sqrt{\frac{g a^2 (4a^2 + 9h^2)}{4ah}} \]

### Final Answer:
\[ u = \frac{1}{2} \sqrt{\frac{g (4a^2 + 9h^2)}{h}} \]