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)=16h_(1)h_(2)`
B. `R^(2)=8h_(1)h_(2)`
C. `R^(2)=4h_(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 the same speed but at different angles.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) for a projectile launched at an angle \( \theta \) with initial speed \( u \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Since the two particles have the same range, we can denote their range as \( R \).

2. **Using the Sine Double Angle Identity**:
We know that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Therefore, we can rewrite the range as:
\[ R = \frac{2u^2 \sin(\theta) \cos(\theta)}{g} \]

3. **Finding Maximum Height**:
The maximum height \( h \) for a projectile launched at angle \( \theta \) is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
For the first particle, the maximum height is \( h_1 \):
\[ h_1 = \frac{u^2 \sin^2(\theta)}{2g} \]
For the second particle, launched at angle \( 90^\circ - \theta \), the maximum height \( h_2 \) is:
\[ h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2(\theta)}{2g} \]

4. **Relating Range and Heights**:
Since both particles have the same range, we can express \( R \) in terms of \( h_1 \) and \( h_2 \):
\[ R = \frac{2u^2 \sin(\theta) \cos(\theta)}{g} \]
Now substituting \( \sin(\theta) \) and \( \cos(\theta) \) in terms of \( h_1 \) and \( h_2 \):
\[ R^2 = \left(\frac{2u^2 \sin(\theta) \cos(\theta)}{g}\right)^2 \]
Expanding this gives:
\[ R^2 = \frac{4u^4 \sin^2(\theta) \cos^2(\theta)}{g^2} \]

5. **Substituting Heights**:
From the height equations:
\[ h_1 = \frac{u^2 \sin^2(\theta)}{2g} \quad \text{and} \quad h_2 = \frac{u^2 \cos^2(\theta)}{2g} \]
We can express \( \sin^2(\theta) \) and \( \cos^2(\theta) \) in terms of \( h_1 \) and \( h_2 \):
\[ \sin^2(\theta) = \frac{2gh_1}{u^2} \quad \text{and} \quad \cos^2(\theta) = \frac{2gh_2}{u^2} \]

6. **Final Relation**:
Substituting these into the equation for \( R^2 \):
\[ R^2 = \frac{4u^4 \cdot \frac{2gh_1}{u^2} \cdot \frac{2gh_2}{u^2}}{g^2} \]
Simplifying gives:
\[ R^2 = \frac{16g^2h_1h_2}{u^4} \]
Rearranging yields:
\[ R^2 = 16h_1h_2 \]
Thus, the relation between \( R \), \( h_1 \), and \( h_2 \) is:
\[ R^2 = 16h_1h_2 \]

### Final Answer:
The relation between \( R \), \( h_1 \), and \( h_2 \) is:
\[ R^2 = 16h_1h_2 \]