The height of a tower is h meter, A body is thrown from the top of tower vertically upwards with some speed, it takes `t_(1)` seconds to reach the ground. Another body thrown from the top of tower with same speed downwards and takes `t_(2)` seconds to reach the ground. If third body, released from same place takes 't' second to reach the ground, then

Options

A. `t = (t_(1) + t_(2))/(2)`
B. `t = (t_(1))/(t_(2))`
C. `(2)/(t) = (1)/(t_(1)) + (1)/(t_(2))`
D. `t = sqrt((t_(1)t_(2)))`

Explanation

To solve the problem, we need to analyze the motion of three bodies thrown from the top of a tower of height \( h \) meters. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Motion of the First Body (Upward Throw)**:
- A body is thrown vertically upwards with an initial speed \( u \) and takes \( t_1 \) seconds to reach the ground.
- The equation of motion can be written as:
\[ -h = ut_1 - \frac{1}{2}gt_1^2 \]
- Rearranging gives:
\[ h = ut_1 - \frac{1}{2}gt_1^2 \quad \text{(Equation 1)} \]

2. **Understanding the Motion of the Second Body (Downward Throw)**:
- Another body is thrown downwards with the same initial speed \( u \) and takes \( t_2 \) seconds to reach the ground.
- The equation of motion for this body is:
\[ -h = ut_2 + \frac{1}{2}gt_2^2 \]
- Rearranging gives:
\[ h = ut_2 + \frac{1}{2}gt_2^2 \quad \text{(Equation 2)} \]

3. **Understanding the Motion of the Third Body (Released)**:
- The third body is simply released (not thrown) and takes \( t \) seconds to reach the ground.
- The equation of motion for this body is:
\[ -h = 0 \cdot t + \frac{1}{2}gt^2 \]
- Rearranging gives:
\[ h = \frac{1}{2}gt^2 \quad \text{(Equation 3)} \]

4. **Equating the Expressions for Height \( h \)**:
- From Equations 1 and 2, we can equate the two expressions for \( h \):
\[ ut_1 - \frac{1}{2}gt_1^2 = ut_2 + \frac{1}{2}gt_2^2 \]
- Rearranging gives:
\[ u(t_1 - t_2) = \frac{1}{2}g(t_2^2 + t_1^2) \quad \text{(Equation 4)} \]

5. **Substituting \( h \) from Equation 3 into Equation 4**:
- We can substitute \( h = \frac{1}{2}gt^2 \) into either Equation 1 or Equation 2 to find a relationship involving \( t \):
\[ \frac{1}{2}gt^2 = ut_1 - \frac{1}{2}gt_1^2 \]
- Rearranging gives:
\[ ut_1 = \frac{1}{2}gt^2 + \frac{1}{2}gt_1^2 \]
- Similarly, substituting into Equation 2 gives:
\[ \frac{1}{2}gt^2 = ut_2 + \frac{1}{2}gt_2^2 \]
- Rearranging gives:
\[ ut_2 = \frac{1}{2}gt^2 - \frac{1}{2}gt_2^2 \]

6. **Finding the Relationship Between \( t, t_1, \) and \( t_2 \)**:
- From the two equations for \( u \):
\[ \frac{ut_1 - ut_2}{t_1 - t_2} = \frac{1}{2}g\left(\frac{t^2 + t_1^2 - t_2^2}{t_1 - t_2}\right) \]
- This implies that:
\[ t = \sqrt{t_1 t_2} \]

### Final Result:
Thus, the time \( t \) taken by the third body released from the same height is given by:
\[ t = \sqrt{t_1 t_2} \]