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 determine the velocity of projection \( U \) of a body that is projected vertically upwards, given the times \( t_1 \) and \( t_2 \) for ascending and descending to a height \( H \) respectively.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When the body is projected upwards, it takes time \( t_1 \) to reach height \( H \).
- After reaching the maximum height, it takes time \( t_2 \) to descend back to the ground from height \( H \).

2. **Using the Equations of Motion**:
- For the upward motion (ascending to height \( H \)):
\[ H = Ut_1 - \frac{1}{2} g t_1^2 \quad \text{(1)} \]
- For the downward motion (descending from height \( H \)):
\[ H = Ut_2 - \frac{1}{2} g t_2^2 \quad \text{(2)} \]

3. **Setting the Two Equations Equal**:
- Since both equations equal \( H \), we can set them equal to each other:
\[ Ut_1 - \frac{1}{2} g t_1^2 = Ut_2 - \frac{1}{2} g t_2^2 \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ Ut_1 - Ut_2 = \frac{1}{2} g t_1^2 - \frac{1}{2} g t_2^2 \]
- Factoring out \( U \) and \( \frac{1}{2} g \):
\[ U(t_1 - t_2) = \frac{1}{2} g (t_1^2 - t_2^2) \]

5. **Using the Difference of Squares**:
- The term \( t_1^2 - t_2^2 \) can be factored using the difference of squares:
\[ t_1^2 - t_2^2 = (t_1 - t_2)(t_1 + t_2) \]
- Substituting this back into the equation gives:
\[ U(t_1 - t_2) = \frac{1}{2} g (t_1 - t_2)(t_1 + t_2) \]

6. **Cancelling Common Terms**:
- Assuming \( t_1 \neq t_2 \), we can cancel \( (t_1 - t_2) \) from both sides:
\[ U = \frac{1}{2} g (t_1 + t_2) \]

7. **Final Result**:
- Thus, the velocity of projection \( U \) is:
\[ U = \frac{g}{2} (t_1 + t_2) \]