D

Step 1: Understanding the scenario
Three particles A, B, and C are thrown with the same initial speed from the top of a tower:
• A is thrown upwards
• B is thrown downwards
• C is thrown horizontally.

In free fall under gravity, the vertical component of velocity increases for both A (after reaching the highest point) and B, while C maintains its horizontal velocity.

Step 2: Analyzing their final speeds
Using the kinematic equations:
1. For particle A, when thrown upwards, it will first decelerate until it reaches the maximum height, after which it will fall back down, gaining speed due to gravity.
2. For particle B, it is thrown downwards; its velocity will increase from the start due to gravity.
3. For particle C, it only has its horizontal component (initial speed) and the vertical component is affected only by gravitational acceleration.

Step 3: Final speeds at ground impact
Considering the equations for final velocity:
• The speed of particle A when it hits the ground will be greater than its initial speed because of the additional potential energy converted to kinetic energy during falling.
• The speed of particle B will also exceed its initial speed due to both its initial downward velocity and gravitational acceleration.
• The horizontal velocity of C at ground level will remain unchanged in the horizontal direction, but the vertical speed will also increase due to free fall.

Upon calculating, we find:
• Final speed of A: $v_A = ext{initial speed} + ext{gain from falling distance}$
• Final speed of B: $v_B > ext{initial speed} + ext{gain from falling distance}$
• Final speed of C: $v_C < v_A, v_B$ (because the initial upward and downward velocities for A and B are greater and gain from gravitational acceleration).

Step 4: Comparing speeds
This analysis shows that:
• $v_B > v_A > v_C$
Hence, Option D: $v_B > v_C > v_A$ is the correct answer.