This question involves projectile motion. The maximum height a stone can reach when thrown vertically gives us information about the initial velocity. Then, we use that to find the maximum horizontal range when thrown at an angle for maximum range.

Step 1: Find the initial velocity (u)

When the stone is thrown straight up (vertically), the maximum height (H) is given by the formula:

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

We are given that H = 10 m. Therefore, we can write:

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

Solving for u²:

$u^2=20g$

We will keep it as u² = 20g for now.

Step 2: Find the maximum horizontal range (R_max)

The formula for the range (R) of a projectile launched with velocity u at an angle θ is:

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

The range is maximum when sin(2θ) is maximum. The maximum value of sin(2θ) is 1, which occurs when 2θ = 90°, or θ = 45°.

Therefore, the formula for the maximum range is:

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

Step 3: Substitute the value of u²

From Step 1, we found that u² = 20g. Let's substitute this into the maximum range formula:

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

The 'g' cancels out, leaving us with R_max = 20 meters.

Final Answer: The maximum horizontal distance the boy can throw the stone is 20 m.

Related Topics & Formulae

Projectile Motion: The motion of an object thrown into the air, subject to gravity. Its path is a parabola.

Key Formulae:

• Maximum Height: $H=\frac{u^2}{2g}{\left(\sin \theta \right)}^2$
• Time of Flight: $T=\frac{2u\sin \theta }{g}$
• Range: $R=\frac{u^2}{g}\sin 2\theta$
• Maximum Range (at θ=45°): $R_{max}=\frac{u^2}{g}$