91Ó°ÊÓ

We will first calculate the initial velocity \(v_{0}\) when the stone reaches the maximum height. We will use the following kinematic equation: \[v^2 = v_{0}^2 - 2g\Delta y\] At the maximum height, the stone has zero vertical velocity (\(v=0\)). The acceleration due to gravity is \(g = 9.81 \mathrm{~m/s^2}\) (approximated to \(10 \mathrm{~m/s^2}\) for simplicity), and the maximum height is \(\Delta y = 10 \mathrm{~m}\).

Plugging the values from step 1 into the equation, we can solve for \(v_{0}\): \[0^2 = v_{0}^2 - 2(10 \mathrm{~m/s^2})(10 \mathrm{~m})\] Solving for \(v_{0}\): \[v_{0}^2 = 20(10 \mathrm{~m})\] \[v_{0} = \sqrt{20(10 \mathrm{~m})} = 10 \sqrt{2} \mathrm{~m/s}\]

For the stone to travel the maximum horizontal distance, it must be thrown at a 45-degree angle to the ground. This is because the projectile's range is maximized when its initial velocity is split into equal horizontal and vertical components. Therefore, we can use the 45-degree angle to determine the horizontal component of the initial velocity, which is: \[v_{0,x} = v_{0} \cos 45^\circ = 10 \sqrt{2} \mathrm{~m/s} \times \frac{1}{\sqrt{2}} = 10 \mathrm{~m/s}\]

Finally, we can use the horizontal component of the initial velocity and the time of flight to calculate the maximum horizontal distance (\(R\)): \[R = v_{0,x} \times t\] Since the stone reaches the maximum height in half of the total flight time, we double the time of flight to get the total flight time (\(t\)), using the max height equation from step 1: \[\Delta y = v_{0}t - \frac{1}{2}gt^2\] \[10 \mathrm{~m} = (10 \sqrt{2} \mathrm{~m/s})\frac{t}{2} - \frac{1}{2}(10 \mathrm{~m/s^2})\frac{t^2}{2}\] Solving for \(t\), we get \(t = 2 \sqrt{2} \mathrm{~s}\). Plugging this value into the range equation from step 3, we get: \[R = (10 \mathrm{~m/s})(2 \sqrt{2} \mathrm{~s}) = 20 \sqrt{2} \mathrm{~m}\] So the maximum horizontal distance the boy can throw the stone up to is: \[\boxed{20 \sqrt{2} \mathrm{~m}}\] which corresponds to option (A).