A body projected vertically upwards from A, the top of a tower, reaches the ground in `t_1` seconds.If it is projected vertically downwards from A with the same velocity, it reaches the ground in `t_2` seconds.If it falls freely from rest at A, in what time does it reach the ground ?

Answer

To solve the problem, we need to analyze the three scenarios in which a body is projected from the top of a tower. We will denote the height of the tower as \( H \), the initial velocity as \( V \), the acceleration due to gravity as \( g \), and the times taken to reach the ground as \( t_1 \) for upward projection, \( t_2 \) for downward projection, and \( t_0 \) for free fall.

### Step-by-Step Solution:

1. **Equation for upward projection**:
When the body is projected upwards with velocity \( V \), it will first rise to a maximum height and then fall back down to the ground. The equation of motion for this scenario is:
\[ -H = V t_1 - \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ H = V t_1 - \frac{1}{2} g t_1^2 \quad \text{(1)} \]

2. **Equation for downward projection**:
When the body is projected downwards with the same velocity \( V \), the equation of motion is:
\[ -H = -V t_2 - \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ H = V t_2 + \frac{1}{2} g t_2^2 \quad \text{(2)} \]

3. **Equating the two equations**:
From equations (1) and (2), we can set them equal to each other:
\[ V t_1 - \frac{1}{2} g t_1^2 = V t_2 + \frac{1}{2} g t_2^2 \]
Rearranging yields:
\[ V t_1 - V t_2 = \frac{1}{2} g t_1^2 + \frac{1}{2} g t_2^2 \]
Factoring out \( V \) gives:
\[ V(t_1 - t_2) = \frac{1}{2} g (t_1^2 + t_2^2) \quad \text{(3)} \]

4. **Finding the relationship between \( H \), \( t_1 \), and \( t_2 \)**:
From equations (1) and (2), we can derive:
\[ H = \frac{1}{2} g t_1 t_2 \]
This can be derived by adding the two equations:
\[ V(t_1 + t_2) = \frac{1}{2} g (t_1 + t_2) \]
Thus, we can express \( H \) as:
\[ H = \frac{g}{2} t_1 t_2 \quad \text{(4)} \]

5. **Finding the time for free fall \( t_0 \)**:
For free fall from rest, we can use the equation:
\[ H = \frac{1}{2} g t_0^2 \]
Substituting \( H \) from equation (4):
\[ \frac{g}{2} t_1 t_2 = \frac{1}{2} g t_0^2 \]
Canceling \( \frac{1}{2} g \) from both sides gives:
\[ t_1 t_2 = t_0^2 \]
Taking the square root of both sides, we find:
\[ t_0 = \sqrt{t_1 t_2} \]

### Final Answer:
The time taken for the body to reach the ground when dropped freely from the top of the tower is:
\[ t_0 = \sqrt{t_1 t_2} \]