B

To solve this problem, we analyze the motion of the three particles A, B, and C thrown from a tower. All three are thrown with the same initial speed but in different directions: A is thrown upwards, B is thrown downwards, and C is thrown horizontally.
1. Velocity of Particle A (upwards): As it is thrown upwards, it'll decelerate due to gravity until it reaches its maximum height, where its velocity becomes zero before falling down. Therefore, while falling back down, its final speed will be influenced by the height and will equal that of B and C when hitting the ground.
2. Velocity of Particle B (downwards): This particle is thrown straight down. The speed it hits the ground will be the initial speed plus any additional speed from gravitational acceleration, increasing the total fall speed.
3. Velocity of Particle C (horizontal): The horizontal velocity does not initially change due to gravity affecting the vertical direction. Hence, it will also hit the ground with a final speed equal to B and A when accounting for the vertical drop of the same height.
Since we want the net speeds upon impact, the particle that was thrown downwards (B) would combine its initial launch speed with the speed accumulated from gravity, making it the fastest. The final impact speeds can be expressed as:
- For A: $$v_A = ext{initial speed} + ext{impact speed from fall}$$.
- For B: $$v_B = ext{initial speed} + ext{(height drop effect)}$$.
- For C: Similar contributions can be calculated, but with variations that yield B as the fastest impact.
Thus the correct option, representing the particle with the highest impact speed, would be B.