A ball is projected with the velocity an angle of `theta` and (90°-`theta`) with horizontal.If `h_1` and `h_2` are maximum heights attained by it in two paths and R is the range of a projectile, then which of the following relation is correct

Options

A. `R = h_1h_2`
B. `R = h_1+h_2`
C. `R = 4(h_1h_2)^(1/2)`
D. `R = (h_1h_2)^(1/3)`

Answer

To solve the problem, we need to analyze the projectile motion of a ball projected at two angles: \( \theta \) and \( 90^\circ - \theta \). We will derive the maximum heights \( h_1 \) and \( h_2 \) for each angle and relate them to the range \( R \) of the projectile.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is projected at two angles: \( \theta \) and \( 90^\circ - \theta \).
- We need to find the maximum heights \( h_1 \) and \( h_2 \) for these angles and relate them to the range \( R \).

2. **Formulas for Maximum Height**:
- The maximum height \( h \) attained by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the angle \( \theta \):
\[ h_1 = \frac{u^2 \sin^2(\theta)}{2g} \]
- For the angle \( 90^\circ - \theta \):
\[ h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2(\theta)}{2g} \]

3. **Formulas for Range**:
- The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ R = \frac{u^2 \cdot 2 \sin(\theta) \cos(\theta)}{g} \]

4. **Relating Heights to Range**:
- From the expressions for \( h_1 \) and \( h_2 \):
\[ h_1 = \frac{u^2 \sin^2(\theta)}{2g} \quad \text{and} \quad h_2 = \frac{u^2 \cos^2(\theta)}{2g} \]
- We can express \( \sin(\theta) \) and \( \cos(\theta) \) in terms of \( h_1 \) and \( h_2 \):
\[ \sin^2(\theta) = \frac{2gh_1}{u^2} \quad \text{and} \quad \cos^2(\theta) = \frac{2gh_2}{u^2} \]

5. **Substituting into the Range Formula**:
- Substitute \( \sin(\theta) \) and \( \cos(\theta) \) into the range formula:
\[ R = \frac{u^2 \cdot 2 \cdot \sqrt{\frac{2gh_1}{u^2}} \cdot \sqrt{\frac{2gh_2}{u^2}}}{g} \]
- Simplifying this gives:
\[ R = \frac{2u^2 \cdot \sqrt{4gh_1h_2}}{u^2g} = \frac{4\sqrt{h_1h_2}}{g} \]

6. **Final Relation**:
- The final relation we derive is:
\[ R = 4\sqrt{h_1h_2} \]

### Conclusion:
The correct relation among the maximum heights \( h_1 \), \( h_2 \), and the range \( R \) is:
\[ R = 4\sqrt{h_1h_2} \]