The maximum horizontal distance that the boy can throw the stone up to will be achieved when the stone is thrown at an angle of 45 degrees with the horizontal. At this angle, the horizontal and vertical components of the stone's velocity will be equal.

Let's assume the initial velocity of the stone is v.
The vertical component of the velocity at the maximum height is 0 m/s, as the stone momentarily stops moving vertically.
Using the kinematic equation for vertical motion, we can find the time it takes for the stone to reach the maximum height:

v = u + at
0 = v + (-9.8)t (taking acceleration due to gravity as -9.8 m/s^2)
t = v/9.8

Using the same equation, we can find the time it takes for the stone to fall back to the ground:

0 = 0 + (-9.8)t'
t' = 0/9.8 = 0

The total time of flight, T, is the sum of the time to reach the maximum height and the time to fall back to the ground:

T = t + t' = v/9.8 + 0 = v/9.8

The horizontal distance, d, traveled by the stone can be calculated using the horizontal component of the velocity and the time of flight:

d = (v * cosθ) * T = v * cosθ * (v/9.8)

As we want to find the maximum horizontal distance, we need to maximize d. Since cosθ has its maximum value of 1 at 0 degrees, we need to throw the stone at an angle of 0 degrees to maximize the horizontal distance.

Therefore, the maximum horizontal distance that the boy can throw the stone up to will be:

d = v * cos0 * (v/9.8) = v * (v/9.8) = (v^2)/9.8

Given that the maximum height reached by the stone is 10 m, we can use the formula for maximum height in vertical motion:

H = (v^2)/(2g) (taking g = 9.8 m/s^2)
10 = (v^2)/(2*9.8)
v^2 = 10 * 2 * 9.8 = 196
v = √196 = 14 m/s

Substituting the value of v into the expression for the maximum horizontal distance:

d = (14^2)/9.8 = 196/9.8 = 20 m

Therefore, the maximum horizontal distance that the boy can throw the stone up to is 20 meters.