From the top of a tower a body A is thrown up vertically with velocity u and another body B is thrown vertically down with the same velocity u. If `v_(A)` and `V_(B)` are their velocities when they reach the ground and `t_(A) and t_(B)` are their times of flight, then

Options

A. `V_(A) = V_(B) and t_(A) = t_(B)`
B. `V_(A) = V_(B) and t_(A) gt t_(B)`
C. `V_(A) = V_(B) and t_(A) = t_(B)`
D. `V_(A) lt V_(B) and t_(A) lt t_(B)`

Answer

To solve the problem, we need to analyze the motion of two bodies A and B thrown from the top of a tower. Body A is thrown upwards with an initial velocity \( u \), and body B is thrown downwards with the same initial velocity \( u \). We will determine their final velocities when they reach the ground and their times of flight.

### Step-by-Step Solution

1. **Define the Variables:**
- Let \( H \) be the height of the tower.
- Let \( t_A \) be the time of flight for body A (thrown upwards).
- Let \( t_B \) be the time of flight for body B (thrown downwards).
- Let \( v_A \) be the final velocity of body A when it reaches the ground.
- Let \( v_B \) be the final velocity of body B when it reaches the ground.

2. **Equation of Motion for Body A:**
- For body A, which is thrown upwards, the displacement equation can be written as:
\[ -H = u t_A - \frac{1}{2} g t_A^2 \]
- Rearranging gives:
\[ \frac{1}{2} g t_A^2 - u t_A - H = 0 \]
- This is a quadratic equation in \( t_A \).

3. **Equation of Motion for Body B:**
- For body B, which is thrown downwards, the displacement equation is:
\[ -H = u t_B + \frac{1}{2} g t_B^2 \]
- Rearranging gives:
\[ \frac{1}{2} g t_B^2 + u t_B + H = 0 \]
- This is also a quadratic equation in \( t_B \).

4. **Relating \( t_A \) and \( t_B \):**
- By equating the two equations derived, we can find a relationship between \( t_A \) and \( t_B \).
- After manipulating the equations, we find:
\[ t_A - t_B = \frac{2u}{g} \]
- This implies that \( t_A > t_B \) since \( u \) and \( g \) are positive.

5. **Final Velocity for Body A:**
- Using the equation of motion for body A to find \( v_A \):
\[ v_A^2 = u^2 - 2gH \]
- Here, we take upward direction as positive, hence the negative sign for displacement \( H \).

6. **Final Velocity for Body B:**
- Using the equation of motion for body B:
\[ v_B^2 = u^2 + 2gH \]
- Here, both initial velocity \( u \) and displacement \( H \) are in the same direction (downwards).

7. **Comparing Final Velocities:**
- Since both bodies are thrown with the same initial velocity \( u \) but in opposite directions, we can conclude:
\[ v_A = \sqrt{u^2 - 2gH} \]
\[ v_B = \sqrt{u^2 + 2gH} \]
- Therefore, \( v_A \) and \( v_B \) will not be equal due to the different effects of gravity on their respective motions.

### Conclusion
- The time of flight for body A is greater than that for body B:
\[ t_A > t_B \]
- The final velocities when they reach the ground are:
\[ v_A \text{ and } v_B \text{ are not equal.} \]