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 complementary angles, \(\theta\) and \(90^\circ - \theta\). We will derive the relationships for the maximum heights \(h_1\) and \(h_2\) and the range \(R\) of the projectile.

### Step-by-Step Solution:

1. **Understanding the Range for Both Angles**:
The range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For the angle \(90^\circ - \theta\), we have:
\[ R = \frac{u^2 \sin(2(90^\circ - \theta))}{g} = \frac{u^2 \sin(180^\circ - 2\theta)}{g} = \frac{u^2 \sin(2\theta)}{g} \]
Therefore, the range for both angles is the same:
\[ R_1 = R_2 = R \]

**Hint**: Remember that \(\sin(180^\circ - x) = \sin(x)\).

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

**Hint**: Use the identity \(\sin(90^\circ - x) = \cos(x)\) to find \(h_2\).

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

**Hint**: Remember that the product of \(\sin^2\) and \(\cos^2\) can be simplified using the double angle identity.

4. **Taking the Square Root**:
To find \(\sqrt{h_1 \cdot h_2}\):
\[ \sqrt{h_1 \cdot h_2} = \sqrt{\frac{u^4 \sin^2(\theta) \cos^2(\theta)}{4g^2}} = \frac{u^2 \sin(\theta) \cos(\theta)}{2g} \]

**Hint**: The square root of a product can be taken separately for each term.

5. **Relating to the Range**:
We know that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Therefore:
\[ \sqrt{h_1 \cdot h_2} = \frac{u^2 \sin(2\theta)}{4g} \]

6. **Expressing Range in Terms of Height**:
From the range formula, we have:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Thus:
\[ \sqrt{h_1 \cdot h_2} = \frac{R}{4} \]

7. **Final Relation**:
Therefore, we can conclude:
\[ R = 4 \sqrt{h_1 \cdot h_2} \]

### Conclusion:
The correct relation is:
\[ R = 4 \sqrt{h_1 \cdot h_2} \]