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. ` sqt ( 8 gh +g^(2) (Delta t)^(2)) `
B. ` sqrt( 8h +4g^(2) (Delta t)^(2))`
C. ` (1)/(2) sqrt( 8gh +g^(2) ( Delta t)^(2))`
D. None of these

Explanation

To solve the problem of finding the initial velocity \( u \) of a ball thrown vertically upwards that is observed at a height \( h \) twice after a time interval \( \Delta t \), we can follow these steps:

### Step 1: Understand the Motion
The motion of the ball can be described using the equations of motion. When the ball is thrown upwards, it will rise to a maximum height and then fall back down. The height \( h \) can be expressed in terms of time \( t \) using the equation:
\[ h = ut - \frac{1}{2}gt^2 \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time.

### Step 2: Set Up the Equation
Since the ball is observed at height \( h \) at two different times \( t_1 \) and \( t_2 \), we can set up the equation for both instances:
\[ h = ut_1 - \frac{1}{2}gt_1^2 \]
\[ h = ut_2 - \frac{1}{2}gt_2^2 \]
Subtracting these two equations gives:
\[ ut_1 - \frac{1}{2}gt_1^2 = ut_2 - \frac{1}{2}gt_2^2 \]

### Step 3: Rearranging the Equation
Rearranging the above equation leads to:
\[ u(t_1 - t_2) = \frac{1}{2}g(t_2^2 - t_1^2) \]
Factoring the right-hand side:
\[ u(t_1 - t_2) = \frac{1}{2}g(t_2 - t_1)(t_2 + t_1) \]

### Step 4: Expressing Time Difference
Let \( \Delta t = t_2 - t_1 \). Thus, we can rewrite the equation as:
\[ u = \frac{g(t_2 + t_1)}{2} \]
Now, we need to find \( t_2 + t_1 \).

### Step 5: Using the Quadratic Formula
From the original height equation, we can express it in standard quadratic form:
\[ \frac{1}{2}gt^2 - ut + h = 0 \]
Using the quadratic formula, the roots \( t_1 \) and \( t_2 \) can be found as:
\[ t = \frac{-(-u) \pm \sqrt{(-u)^2 - 4 \cdot \frac{1}{2}g \cdot h}}{2 \cdot \frac{1}{2}g} \]
This simplifies to:
\[ t = \frac{u \pm \sqrt{u^2 - 2gh}}{g} \]

### Step 6: Finding \( t_1 + t_2 \)
The sum of the roots \( t_1 + t_2 \) is given by:
\[ t_1 + t_2 = \frac{2u}{g} \]

### Step 7: Substitute Back to Find \( u \)
Substituting \( t_1 + t_2 \) back into the equation for \( u \):
\[ u = \frac{g \cdot \frac{2u}{g}}{2} = u \]
This confirms our expression. Now, we can relate \( \Delta t \) to \( u \):
\[ \Delta t = t_2 - t_1 = \sqrt{\frac{2h}{g}} \implies u = \frac{g \Delta t}{2} + \frac{8h}{g \Delta t} \]

### Final Expression for Initial Velocity
Thus, the final expression for the initial velocity \( u \) is:
\[ u = \frac{1}{2} \sqrt{g \left( \Delta t^2 + \frac{8h}{g} \right)} \]