A person can throw a stone to a maximum distance of `h` meter. The greatest height to which he can throw the stone is:

Options

A. `h`
B. `h//2`
C. `2h`
D. `3h`

Explanation

To solve the problem, we need to determine the maximum height to which a person can throw a stone, given that the maximum horizontal distance (range) is \( h \) meters.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Maximizing the Range**:
The range is maximized when \( \sin(2\theta) \) is at its maximum value, which is 1. This occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).
Thus, the maximum range can be expressed as:
\[ R_{\text{max}} = \frac{u^2}{g} \]
Given that the maximum range \( R = h \), we can equate:
\[ h = \frac{u^2}{g} \]

3. **Finding the Initial Velocity**:
Rearranging the equation for \( u^2 \):
\[ u^2 = hg \]

4. **Finding the Maximum Height**:
The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
For maximum height, we use \( \theta = 90^\circ \) (throwing straight up), where \( \sin^2(90^\circ) = 1 \):
\[ H = \frac{u^2}{2g} \]

5. **Substituting for \( u^2 \)**:
Substitute \( u^2 = hg \) into the height formula:
\[ H = \frac{hg}{2g} = \frac{h}{2} \]

### Final Answer:
The greatest height to which the person can throw the stone is:
\[ H = \frac{h}{2} \]