For given value of u, there are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection.

Explanation

To solve the problem, we need to show that the sum of the maximum heights for two angles of projection that give the same horizontal range is independent of the angle of projection. Let's denote the initial velocity as \( u \), the angle of projection for the first case as \( \alpha \), and for the second case as \( \beta \).

### Step-by-Step Solution:

1. **Understanding the Horizontal Range**:
The horizontal range \( R \) for a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Setting Up the Equations**:
For angles \( \alpha \) and \( \beta \), since the ranges are equal, we can write:
\[ \frac{u^2 \sin 2\alpha}{g} = \frac{u^2 \sin 2\beta}{g} \]
Simplifying this, we have:
\[ \sin 2\alpha = \sin 2\beta \]

3. **Using the Sine Identity**:
The sine function has the property that \( \sin x = \sin(180^\circ - x) \). Therefore, we can write:
\[ 2\alpha = 180^\circ - 2\beta \quad \text{or} \quad 2\alpha = 2\beta \]
From the first equation, we can simplify to:
\[ \alpha + \beta = 90^\circ \]
This implies:
\[ \alpha = 90^\circ - \beta \]

4. **Calculating Maximum Heights**:
The maximum height \( h \) for a projectile launched at angle \( \theta \) is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
Therefore, for angles \( \alpha \) and \( \beta \):
\[ h_1 = \frac{u^2 \sin^2 \alpha}{2g} \quad \text{and} \quad h_2 = \frac{u^2 \sin^2 \beta}{2g} \]

5. **Finding the Sum of Maximum Heights**:
Now, we need to find the sum of the maximum heights:
\[ h_1 + h_2 = \frac{u^2 \sin^2 \alpha}{2g} + \frac{u^2 \sin^2 \beta}{2g} \]
Factoring out \( \frac{u^2}{2g} \):
\[ h_1 + h_2 = \frac{u^2}{2g} (\sin^2 \alpha + \sin^2 \beta) \]

6. **Using the Angle Relationship**:
Since \( \alpha = 90^\circ - \beta \), we have:
\[ \sin \alpha = \cos \beta \]
Therefore:
\[ \sin^2 \alpha + \sin^2 \beta = \cos^2 \beta + \sin^2 \beta = 1 \]

7. **Final Result**:
Substituting this back into our equation for the sum of heights:
\[ h_1 + h_2 = \frac{u^2}{2g} \cdot 1 = \frac{u^2}{2g} \]
This shows that the sum of the maximum heights \( h_1 + h_2 \) is independent of the angles \( \alpha \) and \( \beta \).

### Conclusion:
Thus, we have shown that the sum of the maximum heights for the two angles of projection that give the same horizontal range is independent of the angle of projection.