To determine how high a person can throw a ball given the maximum range, we need to use the principles of projectile motion in physics. The maximum range of a projectile is given by the formula:

$$ R = \frac{{v_0^2 \sin(2\theta)}}{g} $$

where:

• $$ R $$ is the range
• $$ v_0 $$ is the initial velocity
• $$ \theta $$ is the angle of projection
• $$ g $$ is the acceleration due to gravity (approximately $$ 9.8 \, \text{m/s}^2 $$)

The maximum range is achieved when $$ \theta = 45^\circ $$, thus $$ \sin(2\theta) = \sin(90^\circ) = 1 $$.

Given that the maximum range $$ R = 100 \, \text{m} $$, we can rewrite the range formula as:

$$ 100 = \frac{{v_0^2 \cdot 1}}{9.8} $$

Solving for $$ v_0^2 $$:

$$ v_0^2 = 100 \times 9.8 = 980 $$

Next, the maximum height $$ H $$ reached by the ball can be calculated using the vertical component of the velocity. The formula for the maximum height is:

$$ H = \frac{{v_0^2 \sin^2(\theta)}}{2g} $$

Here, for a vertical throw, $$ \theta = 90^\circ $$, and thus $$ \sin(90^\circ) = 1 $$.

Substituting the values, we get:

$$ H = \frac{{v_0^2 \cdot 1}}{2 \cdot 9.8} $$

Using $$ v_0^2 = 980 $$, we have:

$$ H = \frac{980}{2 \times 9.8} = \frac{980}{19.6} = 50 \, \text{m} $$

Therefore, the correct answer is:

Option B: 50 m