If `h_(1)` and `h_(2)` be the greatest heights in the two paths of a projectile and a given horizontal range R, then `R=4h_(1) h_(2)`.

Answer

To solve the problem, we need to establish the relationship between the greatest heights \( h_1 \) and \( h_2 \) of two projectile paths that have the same horizontal range \( R \).

### Step-by-Step Solution:

1. **Understanding the Projectile Motion**:
When a projectile is launched at an angle \( \theta \) with an initial velocity \( u \), it follows a parabolic trajectory. The horizontal range \( R \) and the maximum height \( h \) can be derived from the equations of motion.

2. **Formula for Range**:
The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

3. **Formula for Maximum Height**:
The maximum height \( h \) reached by the projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]

4. **Two Angles**:
For two different angles \( \theta \) and \( 90^\circ - \theta \), the heights \( h_1 \) and \( h_2 \) can be expressed as:
- For angle \( \theta \):
\[ h_1 = \frac{u^2 \sin^2(\theta)}{2g} \]
- For angle \( 90^\circ - \theta \):
\[ h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2(\theta)}{2g} \]

5. **Relating Heights to Range**:
Now, substituting \( h_1 \) and \( h_2 \) into the equation \( R = 4h_1 h_2 \):
\[ R = 4 \left( \frac{u^2 \sin^2(\theta)}{2g} \right) \left( \frac{u^2 \cos^2(\theta)}{2g} \right) \]
Simplifying this gives:
\[ R = 4 \cdot \frac{u^4 \sin^2(\theta) \cos^2(\theta)}{4g^2} = \frac{u^4 \sin^2(2\theta)}{4g^2} \]

6. **Final Verification**:
Since we have established that the range \( R \) can be expressed in terms of \( h_1 \) and \( h_2 \) as \( R = 4h_1 h_2 \), we conclude that the statement is true.

### Conclusion:
Thus, the relationship \( R = 4h_1 h_2 \) holds true for the projectile motion under the given conditions.