From the top of a tower 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)gtV_(B)` and `t_(A)gtt_(B)`
C. `V_(A)=V_(B)` and `t_(A)gtt_(B)`
D. `V_(A)ltV_(B)` and `t_(A)ltt_(B)`

Answer

To solve the problem, we need to analyze the motion of both bodies A and B, which are thrown from the top of a tower. Let's denote the height of the tower as \( h \).

### Step 1: Analyze the motion of body A (thrown upwards)
- Body A is thrown upwards with an initial velocity \( u \).
- The time taken by body A to reach the ground is \( t_A \).
- As it goes up, it will first come to a stop and then fall back down to the ground.

Using the kinematic equation for the upward motion:
\[ v_A^2 = u^2 - 2gh \]
where \( v_A \) is the final velocity when it reaches the ground.

### Step 2: Analyze the motion of body B (thrown downwards)
- Body B is thrown downwards with the same initial velocity \( u \).
- The time taken by body B to reach the ground is \( t_B \).

Using the kinematic equation for the downward motion:
\[ v_B^2 = u^2 + 2gh \]
where \( v_B \) is the final velocity when it reaches the ground.

### Step 3: Relate the velocities of A and B
From the equations derived:
1. For body A:
\[ v_A^2 = u^2 - 2gh \]
2. For body B:
\[ v_B^2 = u^2 + 2gh \]

### Step 4: Compare the velocities
Since both bodies are thrown with the same initial velocity \( u \), we can compare their final velocities:
- The final velocity of body A when it reaches the ground is:
\[ v_A = \sqrt{u^2 - 2gh} \]
- The final velocity of body B when it reaches the ground is:
\[ v_B = \sqrt{u^2 + 2gh} \]

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

### Step 5: Analyze the times of flight
For body A, the total time of flight \( t_A \) can be expressed as:
\[ t_A = t_{up} + t_{down} \]
- The time to reach the maximum height \( t_{up} \) is given by:
\[ t_{up} = \frac{u}{g} \]
- The time to fall back down from the maximum height to the ground can be calculated using:
\[ h + \frac{u^2}{2g} = \frac{1}{2} g t_{down}^2 \implies t_{down} = \sqrt{\frac{2h + \frac{u^2}{g}}{g}} \]

For body B, the time of flight \( t_B \) is:
\[ t_B = \sqrt{\frac{2h}{g}} + \frac{u}{g} \]

### Step 6: Compare the times of flight
From the analysis, we can conclude that:
\[ t_A > t_B \]

### Conclusion
Thus, we find that:
- \( v_A < v_B \)
- \( t_A > t_B \)

### Final Answer
- \( v_A = v_B \) (not true)
- \( t_A > t_B \) (true)
- The correct option is that the velocity of body A is less than that of body B, and the time of flight of body A is greater than that of body B.