A body is projected veritclaly upwards.The times corresponding to height h while ascending and while descending are `t_(1)` and `t_(2)` respectively.
Then, the velocity of projection will be (take g as acceleration due to gravity)

Options

A. `gsqrt(t_(1)t_(2))/(2)`
B. `g(t_(1)+t_(2))/(2)`
C. `gsqrt(t_(1)t_(2))`
D. g`(t_(1)t_(2))/((t_(1)+t_(2))`

Answer

To solve the problem, we need to find the initial velocity of a body projected vertically upwards, given the times taken to reach a certain height \( h \) during ascent and descent, denoted as \( t_1 \) and \( t_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a body is projected upwards, it moves against the force of gravity until it reaches its maximum height and then descends back down.
- The time taken to reach a height \( h \) while ascending is \( t_1 \), and the time taken to descend back to the same height \( h \) is \( t_2 \).

2. **Using the Equation of Motion**:
- We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the displacement (height \( h \)), \( u \) is the initial velocity (which we denote as \( u_0 \)), \( a \) is the acceleration (which is \( -g \) when moving upwards), and \( t \) is the time.

3. **Applying the Equation for Ascent**:
- For the ascent to height \( h \):
\[ h = u_0 t_1 - \frac{1}{2} g t_1^2 \]
- Rearranging gives:
\[ 2h = 2u_0 t_1 - g t_1^2 \]

4. **Applying the Equation for Descent**:
- For the descent from height \( h \):
\[ h = u_0 t_2 - \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ 2h = 2u_0 t_2 - g t_2^2 \]

5. **Setting Up the Equations**:
- We now have two equations:
\[ g t_1^2 - 2u_0 t_1 + 2h = 0 \quad \text{(1)} \]
\[ g t_2^2 - 2u_0 t_2 + 2h = 0 \quad \text{(2)} \]

6. **Using the Properties of Quadratic Equations**:
- For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots \( t_1 + t_2 \) is given by \( -\frac{b}{a} \).
- Here, \( a = g \), \( b = -2u_0 \), and \( c = 2h \).
- Thus, we can write:
\[ t_1 + t_2 = \frac{2u_0}{g} \]

7. **Solving for Initial Velocity \( u_0 \)**:
- Rearranging gives:
\[ u_0 = \frac{g(t_1 + t_2)}{2} \]

### Final Answer:
The velocity of projection \( u_0 \) is given by:
\[ u_0 = \frac{g(t_1 + t_2)}{2} \]