1. Find the maximum speed with which the boy can throw the stone

If the boy throws the stone vertically upward, the maximum height reached is given by

$H = \frac{u^2}{2g}$

Given:

$H = 10\text{ m}$

So,

$10 = \frac{u^2}{2g}$

$u^2 = 20g$

2. Find the maximum horizontal distance (range)

For projectile motion on level ground, the horizontal range is

$R = \frac{u^2 \sin 2\theta}{g}$

The range is maximum when

$\sin 2\theta = 1 \quad \Rightarrow \quad \theta = 45^\circ$

Thus,

$R_{\max} = \frac{u^2}{g}$

Substitute $u^2 = 20g$:

$R_{\max} = \frac{20g}{g} = 20\text{ m}$

3. Check options

• A: $20\sqrt{2}$ m — incorrect
• B: 10 m — incorrect
• C: $10\sqrt{2}$ m — incorrect
• D: 20 m — correct

Therefore, the maximum horizontal distance is:

$\boxed{20\text{ m}}$