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. (a) ` 20 sqrt2 ` m
B. (b)` 10 `m
C. (c) ` 10 sqrt2 m `
D. (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 relationship between height and initial velocity**:
The maximum height \( h_{max} \) reached by a projectile is given by the formula:
\[ h_{max} = \frac{u^2}{2g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

2. **Rearranging the formula to find initial velocity**:
From the above formula, we can express the initial velocity \( u \) in terms of the maximum height \( h_{max} \):
\[ u^2 = 2gh_{max} \]
Substituting \( h_{max} = 10 \, \text{m} \):
\[ u^2 = 2 \cdot 9.81 \cdot 10 \]
\[ u^2 = 196.2 \]
\[ u = \sqrt{196.2} \approx 14.0 \, \text{m/s} \]

3. **Using the initial velocity to find the range**:
The horizontal range \( R \) for a projectile launched at an angle \( \theta \) is given by:
\[ R = \frac{u^2}{g} \]
Since we want to find the maximum horizontal distance, we will consider the angle \( \theta = 45^\circ \) for maximum range.

4. **Calculating the range**:
Substituting \( u^2 \) into the range formula:
\[ R = \frac{u^2}{g} = \frac{196.2}{9.81} \]
\[ R \approx 20.0 \, \text{m} \]

5. **Conclusion**:
Therefore, the maximum horizontal distance that the boy can throw the stone is approximately **20 meters**.

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

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