If a person can throw a stone to maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

Options

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

Explanation

To solve the problem, we need to find the maximum horizontal distance (range) a person can throw a stone, given that they can throw it to a maximum height \( h \) meters vertically.

### Step-by-Step Solution:

1. **Understanding Maximum Height**:
The maximum height \( h \) reached by a stone thrown vertically can be expressed using the formula:
\[ h = \frac{u^2}{2g} \]
where \( u \) is the initial velocity of the throw and \( g \) is the acceleration due to gravity.

2. **Rearranging for Initial Velocity**:
From the equation above, we can rearrange it to find the initial velocity \( u \):
\[ u^2 = 2gh \]

3. **Maximum Range in Projectile Motion**:
When the stone is thrown at an angle (for maximum horizontal distance), the range \( R \) can be expressed as:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For maximum range, \( \sin 2\theta \) should be equal to 1, which occurs when \( \theta = 45^\circ \). Thus, the formula simplifies to:
\[ R_{\text{max}} = \frac{u^2}{g} \]

4. **Substituting for \( u^2 \)**:
Now, we substitute \( u^2 \) from step 2 into the range formula:
\[ R_{\text{max}} = \frac{2gh}{g} \]

5. **Simplifying the Expression**:
The \( g \) in the numerator and denominator cancels out:
\[ R_{\text{max}} = 2h \]

### Final Result:
Thus, the maximum distance through which the stone can be thrown horizontally by the same person is:
\[ \boxed{2h} \]