Two particles are projected obliquely from ground with same speed such that their range `'R'` are same but they attain different maximum heights `h_(1)` and `h_(2)` then relation between `R, h_(1)` and `h_(2)` is:

Options

A. `R^(2) = 4 h_(1)h_(2)`
B. `R^(2)=16 h_(1)h_(2)`
C. `R^(2) =2 h_(1)h_(2)`
D. `R^(2) = h_(1)h_(2)`

Answer

To solve the problem, we need to establish the relationship between the range \( R \) and the maximum heights \( h_1 \) and \( h_2 \) of two particles projected at different angles but with the same initial speed and the same range.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Condition for Same Range**:
Since both particles have the same range \( R \) and are projected with the same speed \( u \), we can denote their angles of projection as \( \theta \) and \( \phi \). The condition for the same range implies:
\[ \sin 2\theta = \sin 2\phi \]
This means:
\[ 2\theta + 2\phi = 180^\circ \quad \text{or} \quad \theta + \phi = 90^\circ \]

3. **Maximum Height Formula**:
The maximum height \( h \) reached by a projectile is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
Therefore, for the two particles, we have:
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \quad \text{and} \quad h_2 = \frac{u^2 \sin^2 \phi}{2g} \]

4. **Using the Angle Relationship**:
Since \( \phi = 90^\circ - \theta \), we can express \( \sin \phi \) in terms of \( \theta \):
\[ \sin \phi = \cos \theta \]
Thus, we can write:
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \quad \text{and} \quad h_2 = \frac{u^2 \cos^2 \theta}{2g} \]

5. **Finding the Product of Heights**:
Now, we can find the product of the two heights:
\[ h_1 \cdot h_2 = \left(\frac{u^2 \sin^2 \theta}{2g}\right) \cdot \left(\frac{u^2 \cos^2 \theta}{2g}\right) = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2} \]

6. **Using the Identity for Sine**:
We know that:
\[ \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta \]
Therefore, we can substitute this into our equation:
\[ h_1 \cdot h_2 = \frac{u^4}{4g^2} \cdot \frac{1}{4} \sin^2 2\theta = \frac{u^4 \sin^2 2\theta}{16g^2} \]

7. **Relating to the Range**:
From the range formula, we have:
\[ R = \frac{u^2 \sin 2\theta}{g} \implies R^2 = \frac{u^4 \sin^2 2\theta}{g^2} \]
Thus, we can express \( \sin^2 2\theta \) in terms of \( R \):
\[ \sin^2 2\theta = \frac{R^2 g^2}{u^4} \]

8. **Final Relationship**:
Substituting this back into the height product equation gives:
\[ h_1 \cdot h_2 = \frac{u^4}{16g^2} \cdot \frac{R^2 g^2}{u^4} = \frac{R^2}{16} \]
Therefore, we arrive at the final relationship:
\[ R^2 = 16 h_1 h_2 \]

### Conclusion:
The relationship between the range \( R \) and the maximum heights \( h_1 \) and \( h_2 \) is:
\[ R^2 = 16 h_1 h_2 \]