1. Use the formula for maximum horizontal range

For a projectile thrown with speed $u$, the horizontal range is

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

The maximum range occurs when

$\theta = 45^\circ$

so

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

Given:

$R_{\max} = 100\ \text{m}$

Hence,

$\frac{u^2}{g} = 100$

So,

$u^2 = 100g$

2. Now find the maximum height the person can throw

To throw the ball to the greatest height above the ground, it should be thrown vertically upward with the same speed $u$.

The maximum height reached is

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

Substitute $u^2 = 100g$:

$H = \frac{100g}{2g} = 50\ \text{m}$

3. Check options

• A: $25\ \text{m}$
• B: $50\ \text{m}$ ✅
• C: $100\ \text{m}$
• D: $200\ \text{m}$

Therefore, the correct answer is:

$\boxed{50\ \text{m}}$

So, Option B is correct.