B

To find the coordinates where the velocity of the projectile is minimum, we can analyze the projectile's motion under ideal conditions (neglecting air resistance).

Step 1: The maximum horizontal range of a projectile is given by the formula:
$$ R = \frac{v^2 \sin(2\theta)}{g} $$
where $R$ is the range, $v$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

Step 2: For maximum range, the angle of projection $\theta$ is $45^\circ$. Thus, we can use $R = 400 m$ to find the required initial velocity. However, since we are concerned with the position of minimum velocity, we need to consider the kinematics of projectile motion.

Step 3: The maximum height $H$ reached by the projectile can be calculated using the formula for range at angle $45^\circ$, which results in:
$$ H = \frac{R}{4} = \frac{400}{4} = 100 m $$.
This implies that the projectile reaches its maximum height of 100 m at half the horizontal range (200 m).

Step 4: When the projectile is at its maximum height, its vertical velocity component is zero, thus the overall velocity is at a minimum value (as it has only horizontal velocity and no vertical component).

Step 5: Therefore, the coordinates at this point where velocity is minimum are (200, 100).
Thus, the correct answer is option B.