From the top of a tower two bodies are projected with the same intital speed of 40 `ms^(-1)` ,first body vertically upwards and second body vertically downwards .A third body is freely released from the top of the tower.If their respective times of flights are `T_(1),T_(2)` and `T_(3)` Identify the correct descending order of the times of flights

Options

A. `T_1,T_2,T_3`
B. `T_2,T_3,T_1`
C. `T_2,T_1,T_3`
D. `T_1,T_3,T_2`

Answer

To solve the problem, we need to analyze the motion of the three bodies projected from the top of the tower. Let's denote the height of the tower as \( h \) and the initial speed of the bodies as \( u = 40 \, \text{m/s} \).

### Step 1: Analyze the first body (T1)
The first body is projected vertically upwards with an initial speed of \( 40 \, \text{m/s} \). It will rise until it reaches its maximum height and then fall back down to the ground.

1. **Time to reach maximum height (upward motion)**:
\[ v = u - gt \implies 0 = 40 - 9.8t_1 \implies t_1 = \frac{40}{9.8} \approx 4.08 \, \text{s} \]
(where \( g \approx 9.8 \, \text{m/s}^2 \))

2. **Height reached (h')**:
\[ h' = ut_1 - \frac{1}{2}gt_1^2 \approx 40 \times 4.08 - \frac{1}{2} \times 9.8 \times (4.08)^2 \approx 81.6 - 81.6 \approx 0 \, \text{m} \]

3. **Total time of flight (T1)**:
The time taken to come down from maximum height back to the ground will be equal to the time taken to go up.
\[ T_1 = t_1 + t_1 = 2t_1 \approx 2 \times 4.08 \approx 8.16 \, \text{s} \]

### Step 2: Analyze the second body (T2)
The second body is projected vertically downwards with the same initial speed of \( 40 \, \text{m/s} \).

1. **Using the second equation of motion**:
\[ h = ut_2 + \frac{1}{2}gt_2^2 \]
Rearranging gives us a quadratic equation in \( t_2 \):
\[ 0 = \frac{1}{2}gt_2^2 + ut_2 - h \]

Solving this quadratic equation using the quadratic formula:
\[ t_2 = \frac{-u \pm \sqrt{u^2 + 2gh}}{g} \]
Since we are looking for positive time, we take the positive root:
\[ t_2 = \frac{-40 + \sqrt{40^2 + 2 \times 9.8 \times h}}{9.8} \]

Since \( h \) is constant, \( T_2 \) will be less than \( T_1 \).

### Step 3: Analyze the third body (T3)
The third body is simply dropped from the top of the tower (initial speed \( u = 0 \)).

1. **Using the second equation of motion**:
\[ h = \frac{1}{2}gt_3^2 \]
Rearranging gives:
\[ t_3 = \sqrt{\frac{2h}{g}} \]
This time will be less than \( T_1 \) but more than \( T_2 \).

### Conclusion: Order of times of flight
Now we can summarize the times of flight:
- \( T_1 \) (upward) is the longest.
- \( T_2 \) (downward) is the shortest.
- \( T_3 \) (freely falling) is in between.

Thus, the correct descending order of the times of flights is:
\[ T_1 > T_3 > T_2 \]