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, we need to find the horizontal range of a projectile that is projected from the ground with a certain speed at an angle θ with the horizontal and at an angle (90° - θ) with the vertical. We will denote the maximum heights attained in these two cases as \( h_1 \) and \( h_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- When the stone is projected at an angle θ with the horizontal, it reaches a maximum height \( h_1 \).
- When projected at an angle (90° - θ) with the vertical, it reaches a maximum height \( h_2 \).

2. **Finding Maximum Heights**:
- The formula for maximum height \( h \) in projectile motion is given by:
\[ h = \frac{u_y^2}{2g} \]
- For the first case (angle θ with horizontal):
\[ h_1 = \frac{(u \sin \theta)^2}{2g} = \frac{u^2 \sin^2 \theta}{2g} \quad \text{(Equation 1)} \]
- For the second case (angle (90° - θ) with vertical):
\[ h_2 = \frac{(u \cos \theta)^2}{2g} = \frac{u^2 \cos^2 \theta}{2g} \quad \text{(Equation 2)} \]

3. **Relating the Heights**:
- From Equations 1 and 2, we can express the product of the heights:
\[ h_1 \cdot h_2 = \left(\frac{u^2 \sin^2 \theta}{2g}\right) \cdot \left(\frac{u^2 \cos^2 \theta}{2g}\right) \]
\[ = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2} \]

4. **Using the Identity**:
- We can use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ h_1 \cdot h_2 = \frac{u^4 \cdot \frac{1}{4} \sin^2 2\theta}{4g^2} = \frac{u^4 \sin^2 2\theta}{16g^2} \]

5. **Finding the Range**:
- The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- Squaring the range:
\[ R^2 = \left(\frac{u^2 \sin 2\theta}{g}\right)^2 = \frac{u^4 \sin^2 2\theta}{g^2} \]

6. **Relating the Range to the Heights**:
- From the earlier relation of heights:
\[ h_1 \cdot h_2 = \frac{R^2}{16} \]
- Thus, we can express the range in terms of the heights:
\[ R^2 = 16 h_1 \cdot h_2 \]
- Taking the square root gives:
\[ R = 4 \sqrt{h_1 \cdot h_2} \]

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