A boy can throw a stone up to a maximum height of `10 m`. The maximum horizontal distance that the boy can throw the same stone up to will be :

Options

A. `20 sqrt(2)m`
B. 10m
C. `10sqrt(2)m`
D. 20m

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution

1. **Understanding the Problem**: We need to find the maximum horizontal distance (range) that a boy can throw a stone, given that he can throw it to a maximum height of 10 meters.

2. **Using the Maximum Height Formula**: The formula for maximum height (h) when a projectile is thrown is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( \theta \) is the angle of projection.

3. **Finding the Initial Velocity for Maximum Height**: For maximum height, the angle \( \theta \) is \( 90^\circ \) (straight up). Thus, \( \sin 90^\circ = 1 \). The equation simplifies to:
\[ h = \frac{u^2}{2g} \]
Given that \( h = 10 \, \text{m} \), we can set up the equation:
\[ 10 = \frac{u^2}{2 \times 10} \]

4. **Solving for \( u^2 \)**:
\[ 10 = \frac{u^2}{20} \]
Multiplying both sides by 20 gives:
\[ u^2 = 200 \]

5. **Using the Range Formula**: The formula for the range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For maximum range, the angle \( \theta \) is \( 45^\circ \). Therefore, \( \sin 2\theta = \sin 90^\circ = 1 \). The range formula simplifies to:
\[ R = \frac{u^2}{g} \]

6. **Substituting Values**: Now we can substitute the values of \( u^2 \) and \( g \):
\[ R = \frac{200}{10} \]

7. **Calculating the Range**:
\[ R = 20 \, \text{m} \]

8. **Conclusion**: The maximum horizontal distance that the boy can throw the stone is \( 20 \, \text{m} \).

### Final Answer
The maximum horizontal distance that the boy can throw the stone is **20 meters**.