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 sqrt2 ` m
B. ` 10 `m
C. ` 10 sqrt2 m `
D. ` 20 m `

Explanation

To solve the problem of finding the maximum horizontal distance a boy can throw a stone given that he can throw it to a maximum height of 10 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Maximum Height**:
The maximum height (h_max) that the boy can throw the stone is given as 10 m. This height is achieved when the stone is thrown vertically upwards.

2. **Use the Kinematic Equation**:
We can use the kinematic equation for vertical motion to relate the initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
\[ v^2 = u^2 + 2as \]
At the maximum height, the final velocity (v) is 0 m/s, the acceleration (a) is -g (where g = 10 m/s²), and the displacement (s) is 10 m.

3. **Set Up the Equation**:
Plugging in the values into the equation:
\[ 0 = u^2 - 2g \cdot h_{max} \]
Rearranging gives:
\[ u^2 = 2gh_{max} \]

4. **Calculate the Initial Velocity**:
Substituting g = 10 m/s² and h_max = 10 m:
\[ u^2 = 2 \cdot 10 \cdot 10 = 200 \]
Therefore, the initial velocity (u) is:
\[ u = \sqrt{200} = 10\sqrt{2} \text{ m/s} \]

5. **Determine the Maximum Range**:
The range (R) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For maximum range, the angle of projection (θ) should be 45 degrees, where sin(90°) = 1. Thus, the formula simplifies to:
\[ R = \frac{u^2}{g} \]

6. **Substitute the Values**:
Now substituting u² = 200 and g = 10:
\[ R = \frac{200}{10} = 20 \text{ m} \]

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