From the top of a tower two bodies are projected with the same initial 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 time 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 \( H \).

### Step 1: Analyze the first body (projected upwards)
The first body is projected vertically upwards with an initial speed \( u = 40 \, \text{m/s} \). The time of flight \( T_1 \) can be calculated using the kinematic equation:
\[ H = u T_1 - \frac{1}{2} g T_1^2 \]
where \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

Rearranging gives:
\[ \frac{1}{2} g T_1^2 - u T_1 + H = 0 \]
This is a quadratic equation in \( T_1 \).

### Step 2: Analyze the second body (projected downwards)
The second body is projected vertically downwards with the same initial speed \( u = 40 \, \text{m/s} \). The time of flight \( T_2 \) can be calculated using the equation:
\[ H = u T_2 + \frac{1}{2} g T_2^2 \]
Rearranging gives:
\[ \frac{1}{2} g T_2^2 + u T_2 - H = 0 \]
This is also a quadratic equation in \( T_2 \).

### Step 3: Analyze the third body (freely falling)
The third body is simply released from rest, so its initial speed \( u = 0 \). The time of flight \( T_3 \) can be calculated using:
\[ H = \frac{1}{2} g T_3^2 \]
Rearranging gives:
\[ T_3^2 = \frac{2H}{g} \implies T_3 = \sqrt{\frac{2H}{g}} \]

### Step 4: Compare the times of flight
Now we need to compare \( T_1 \), \( T_2 \), and \( T_3 \):

1. **For \( T_1 \)**: The body goes up, stops, and then comes down. This means it will take longer than just falling straight down.
2. **For \( T_2 \)**: The body goes down with an initial speed, so it will take less time than \( T_1 \) but more than \( T_3 \).
3. **For \( T_3 \)**: This body falls freely from rest and will take the least time to reach the ground.

### Conclusion
From the analysis, we can conclude:
\[ T_2 > T_1 > T_3 \]
Thus, the correct descending order of the time of flights is:
\[ T_2, T_1, T_3 \]