Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally. They hit the ground with speeds `V_A,V_B and V_C` respectively

Options

A. `V_A = V_B = V_C`
B. `V_A = V_B gt V_C`
C. `V_B gt V_C gt V_A`
D. `V_A gt V_B = V_C`

Explanation

To solve the problem of three particles A, B, and C thrown from the top of a tower with the same speed, we will analyze the motion of each particle and determine their final speeds when they hit the ground.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- Particle A is thrown upwards.
- Particle B is thrown downwards.
- Particle C is thrown horizontally.
- All three particles are thrown with the same initial speed, \( u \).

2. **Analyzing Particle A (Thrown Upwards):**
- When particle A is thrown upwards, it will first move against gravity, slow down, stop momentarily at its highest point, and then fall back down.
- The time taken to reach the highest point can be calculated using the equation:
\[ v = u - gt \]
where \( v = 0 \) at the highest point, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach the highest point.
- Rearranging gives:
\[ t = \frac{u}{g} \]
- The total time taken to reach the ground after being thrown upwards is twice this time (up and down):
\[ t_{A} = 2 \cdot \frac{u}{g} = \frac{2u}{g} \]
- The final velocity \( V_A \) when it hits the ground can be calculated using:
\[ V_A = u + gt_{A} = u + g \cdot \frac{2u}{g} = u + 2u = 3u \]

3. **Analyzing Particle B (Thrown Downwards):**
- Particle B is thrown downwards with the same initial speed \( u \).
- The time taken to hit the ground is the same as for particle A:
\[ t_{B} = \frac{2u}{g} \]
- The final velocity \( V_B \) when it hits the ground can be calculated as:
\[ V_B = u + gt_{B} = u + g \cdot \frac{2u}{g} = u + 2u = 3u \]

4. **Analyzing Particle C (Thrown Horizontally):**
- Particle C is thrown horizontally. The vertical motion is influenced only by gravity.
- The time taken to hit the ground is the same as for particles A and B:
\[ t_{C} = \frac{2u}{g} \]
- The horizontal component of the velocity remains \( u \), and the vertical component can be calculated using:
\[ V_{C_y} = gt_{C} = g \cdot \frac{2u}{g} = 2u \]
- The final speed \( V_C \) when it hits the ground is the resultant of the horizontal and vertical components:
\[ V_C = \sqrt{u^2 + (2u)^2} = \sqrt{u^2 + 4u^2} = \sqrt{5u^2} = u\sqrt{5} \]

5. **Conclusion:**
- The final speeds are:
- \( V_A = 3u \)
- \( V_B = 3u \)
- \( V_C = u\sqrt{5} \)
- Since \( 3u > u\sqrt{5} \), we conclude that:
\[ V_A = V_B > V_C \]