A ball is thrown vertically upwards. It was observed at a height h twice after a time interval `Deltat`. The initial velocity of the ball is

Options

A. `sqrt(8gh+(g Deltat)^(2))`
B. `sqrt(2gh+((gDeltat)/(2))^(2))`
C. `sqrt(8gh+(2gDeltat)^(2))`
D. `sqrt(2gh)`

Explanation

To solve the problem of finding the initial velocity of a ball thrown vertically upwards, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Motion**:
The ball is thrown upwards and reaches a height \( h \) at two different times, \( t_1 \) and \( t_2 \), separated by a time interval \( \Delta t = t_2 - t_1 \).

2. **Using the Second Equation of Motion**:
The second equation of motion states:
\[ S = ut + \frac{1}{2} a t^2 \]
For our case, the displacement \( S \) is \( h \), the initial velocity is \( u \), and the acceleration \( a \) is \( -g \) (since gravity acts downwards).

3. **Setting Up the Equation**:
At height \( h \):
\[ h = ut_1 - \frac{1}{2} g t_1^2 \quad \text{(1)} \]
and
\[ h = ut_2 - \frac{1}{2} g t_2^2 \quad \text{(2)} \]

4. **Equating the Two Equations**:
From equations (1) and (2), we can set them equal to each other:
\[ ut_1 - \frac{1}{2} g t_1^2 = ut_2 - \frac{1}{2} g t_2^2 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ u(t_1 - t_2) = \frac{1}{2} g (t_2^2 - t_1^2) \]
This can be factored as:
\[ u(t_1 - t_2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1) \]

6. **Substituting \( \Delta t \)**:
Let \( \Delta t = t_2 - t_1 \):
\[ u(-\Delta t) = \frac{1}{2} g (\Delta t)(t_2 + t_1) \]
Thus, we can express \( u \):
\[ u = \frac{g(t_2 + t_1)}{2} \]

7. **Finding \( t_2 + t_1 \)**:
From the quadratic equation derived from the motion equations, we know:
\[ t_1 + t_2 = \frac{2u}{g} \]
Therefore:
\[ u = \frac{g \Delta t}{2} + \frac{g \Delta t}{2} = g \frac{\Delta t}{2} \]

8. **Final Expression for Initial Velocity**:
We can also relate \( t_1 t_2 \) to the height \( h \):
\[ t_1 t_2 = \frac{2h}{g} \]
Using these relationships, we can derive:
\[ u^2 = 2gh + \frac{g^2 \Delta t^2}{4} \]
Taking the square root gives:
\[ u = \sqrt{2gh + \frac{g^2 \Delta t^2}{4}} \]

### Final Answer:
Thus, the initial velocity \( u \) of the ball is given by:
\[ u = \sqrt{2gh + \frac{g^2 \Delta t^2}{4}} \]