A body is projected verticallt upwards. If `t_(1)` and `t_(2)` be the times at which it is at a height h above the point of projection while ascending and descending respectively, then:

Options

A. `h = g t_(1)t_(2)`
B. `h = 2 g t_(1)t_(2)`
C. `h = (1)/(2) g t_(1)t_(2)`
D. `h = (1)/(4) g t_(1)t_(2)`

Explanation

To solve the problem, we need to analyze the motion of a body projected vertically upwards. We will derive the relationship between the times \( t_1 \) and \( t_2 \) when the body is at a height \( h \) during its ascent and descent.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a body is projected upwards with an initial velocity \( u \), it will ascend to a maximum height and then descend back down. The times \( t_1 \) and \( t_2 \) correspond to the moments when the body is at height \( h \) during its ascent and descent, respectively.

2. **Using the Equation of Motion**:
- The height \( h \) at any time \( t \) can be expressed using the equation of motion:
\[ h = u t - \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity (acting downwards).

3. **Setting Up the Equation**:
- For the height \( h \) at time \( t_1 \) (ascent):
\[ h = u t_1 - \frac{1}{2} g t_1^2 \]
- For the height \( h \) at time \( t_2 \) (descent):
\[ h = u t_2 - \frac{1}{2} g t_2^2 \]

4. **Rearranging the Equations**:
- Rearranging both equations gives:
\[ \frac{1}{2} g t_1^2 - u t_1 + h = 0 \quad \text{(1)} \]
\[ \frac{1}{2} g t_2^2 - u t_2 + h = 0 \quad \text{(2)} \]

5. **Identifying the Quadratic Equation**:
- Both equations (1) and (2) are quadratic in form. The general form is \( at^2 + bt + c = 0 \).
- For both equations, we can identify:
- \( a = \frac{1}{2} g \)
- \( b = -u \)
- \( c = h \)

6. **Using the Product of Roots**:
- For any quadratic equation \( at^2 + bt + c = 0 \), the product of the roots \( t_1 \) and \( t_2 \) is given by:
\[ t_1 t_2 = \frac{c}{a} \]
- Substituting our values:
\[ t_1 t_2 = \frac{h}{\frac{1}{2} g} = \frac{2h}{g} \]

7. **Final Relationship**:
- Rearranging gives us:
\[ \frac{1}{2} g t_1 t_2 = h \]
- This is the required relationship between the times \( t_1 \) and \( t_2 \) when the body is at height \( h \).

### Conclusion:
The final result is:
\[ \frac{1}{2} g t_1 t_2 = h \]