Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds `v_(A), v_(B)` and `v_(C)` respectively:

Options

A. `v_(A)=v_(B) gt v_(C)`
B. `v_(A)=v_(B)=v_(C)`
C. `v_(A) gt v_(B)=v_(C)`
D. `v_(B) gt v_(C) gt v_(A)`

Explanation

To solve the problem, we need to analyze the motion of the three particles A, B, and C, which are thrown from the top of a tower with the same initial speed \( U \). We will derive the final speeds \( v_A \), \( v_B \), and \( v_C \) for each particle when they hit the ground.

### Step 1: Analyze Particle A (Thrown Upwards)
When particle A is thrown straight up, it will first move upwards, come to a stop at its maximum height, and then fall back down to the ground.

Using the equation of motion:
\[ v_A^2 = U^2 - 2gH \]
At the maximum height, the final velocity \( v_A \) will be zero. When it falls back down to the height \( H \), we can express the final speed as:
\[ v_A^2 = U^2 + 2gH \quad \text{(since it falls back down)} \]

### Step 2: Analyze Particle B (Thrown Downwards)
For particle B, which is thrown straight down, we can use the same equation of motion:
\[ v_B^2 = U^2 + 2gH \]
Since it is thrown downwards, the initial speed adds to the speed gained from falling.

### Step 3: Analyze Particle C (Thrown Horizontally)
For particle C, which is thrown horizontally, we need to consider both horizontal and vertical components of its motion.

1. **Horizontal Component**: The horizontal speed remains constant:
\[ v_{C_x} = U \]

2. **Vertical Component**: The vertical motion is similar to free fall:
\[ v_{C_y}^2 = 0 + 2gH = 2gH \]
Thus, the vertical speed when it hits the ground is:
\[ v_{C_y} = \sqrt{2gH} \]

Now, we can find the resultant speed \( v_C \) using Pythagoras' theorem:
\[ v_C^2 = v_{C_x}^2 + v_{C_y}^2 = U^2 + (2gH) \]

### Step 4: Compare the Speeds
Now we have:
- For particle A:
\[ v_A^2 = U^2 + 2gH \]
- For particle B:
\[ v_B^2 = U^2 + 2gH \]
- For particle C:
\[ v_C^2 = U^2 + 2gH \]

From this, we can conclude that:
\[ v_A = v_B = v_C \]

### Conclusion
All three particles hit the ground with the same speed:
\[ v_A = v_B = v_C \]