A stone projected from ground with certain speed at an angle `theta` with horizontal attains maximum height `h_(1)` when it is projected with same speed at an angle `theta` with vertical attains height `h_(2)`. The horizontal range of projectile is

Options

A. `(h_(1) + h_(2))/(2)`
B. `2h_(1) h_(2)`
C. `4 sqrt(h_(1) h_(2))`
D. `h_(1) + h_(2)`

Answer

To solve the problem of finding the horizontal range of a projectile launched at an angle θ with respect to the horizontal and vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A stone is projected from the ground with an initial speed \( u \) at an angle \( \theta \) with the horizontal, reaching a maximum height \( h_1 \).
- The same stone is projected with the same speed \( u \) at an angle \( \theta \) with the vertical, reaching a maximum height \( h_2 \).

2. **Finding Maximum Heights**:
- For the first case (angle \( \theta \) with horizontal):
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \]
- For the second case (angle \( \theta \) with vertical, which means \( 90^\circ - \theta \) with horizontal):
\[ h_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g} \]

3. **Relating Heights**:
- We can express \( h_1 \) and \( h_2 \) in terms of \( u \):
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \]
\[ h_2 = \frac{u^2 \cos^2 \theta}{2g} \]

4. **Finding the Range**:
- The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- We can express \( u^2 \) in terms of \( h_1 \) and \( h_2 \):
\[ u^2 = 2gh_1 \cdot \frac{1}{\sin^2 \theta} \quad \text{(from \( h_1 \))} \]
\[ u^2 = 2gh_2 \cdot \frac{1}{\cos^2 \theta} \quad \text{(from \( h_2 \))} \]

5. **Multiplying Heights**:
- Multiply \( h_1 \) and \( h_2 \):
\[ h_1 h_2 = \left(\frac{u^2 \sin^2 \theta}{2g}\right) \left(\frac{u^2 \cos^2 \theta}{2g}\right) = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2} \]

6. **Finding the Range in Terms of Heights**:
- From the expression for range:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} \]
- Substitute \( u^2 = 2g \sqrt{h_1 h_2} \):
\[ R = \frac{2g \sqrt{h_1 h_2} \cdot 2 \sin \theta \cos \theta}{g} = 4 \sqrt{h_1 h_2} \sin \theta \cos \theta \]

7. **Final Expression**:
- Since \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can write:
\[ R = 4 \sqrt{h_1 h_2} \]

### Conclusion:
The horizontal range \( R \) of the projectile is given by:
\[ R = 4 \sqrt{h_1 h_2} \]