The two fragments of the shell will fall at a distance of approximately 431 meters from each other.

This can be calculated using the kinematic equations of motion.

The first fragment has a horizontal velocity of 100 m/s and reaches a vertical height of 1960 meters before exploding. At this point, its vertical velocity is equal to 0 since it has reached its maximum height.

Using the equation v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration(acceleration due to gravity is -9.8 m/s^2) and s is displacement, we can find the total time taken by the first fragment to reach its maximum height.

t = sqrt(2(1960)/9.8) = 20 seconds

Now, using the equation s = ut + 1/2at^2, we can find the horizontal distance covered by the first fragment.

s = 100 * 20 = 2000 meters

Therefore, the first fragment will fall on the ground at a distance of 2000 meters horizontally from its starting point.

The second fragment has a horizontal velocity of twice that of the first (200 m/s) and reaches the same maximum height of 1960 meters. Using the equations of motion, we can find that it will take 10 seconds to reach this maximum height and it will fall on the ground at a horizontal distance of 1000 meters from its starting point.

Hence, the total horizontal distance between the two fragments will be equal to the difference between their respective horizontal distances, i.e. 2000 - 1000 = 1000 meters.