A ball is projected with `20sqrt2 m//s` at angle `45^(@)` with horizontal.The angular velocity of the particle at highest point of its journey about point of projection is

Options

A. `0.1 rad//s`
B. `1 rad//s`
C. `0.3 rad//s`
D. `0.4 rad//s`

Explanation

To solve the problem of finding the angular velocity of a ball projected at an angle of 45 degrees with an initial velocity of \(20\sqrt{2} \, \text{m/s}\) at the highest point of its trajectory about the point of projection, we can follow these steps:

### Step 1: Determine the horizontal and vertical components of the initial velocity
The ball is projected at an angle of \(45^\circ\) with an initial velocity of \(20\sqrt{2} \, \text{m/s}\). We can calculate the horizontal (\(V_x\)) and vertical (\(V_y\)) components of the velocity using trigonometric functions:

\[ V_x = V \cdot \cos(\theta) = 20\sqrt{2} \cdot \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

\[ V_y = V \cdot \sin(\theta) = 20\sqrt{2} \cdot \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

### Step 2: Find the velocity at the highest point
At the highest point of the trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged. Therefore, the velocity at the highest point (\(V\)) is:

\[ V = V_x = 20 \, \text{m/s} \]

### Step 3: Calculate the maximum height (H_max)
To find the maximum height reached by the ball, we can use the formula:

\[ H_{\text{max}} = \frac{U^2 \sin^2(\theta)}{2g} \]

Substituting the values:

\[ H_{\text{max}} = \frac{(20\sqrt{2})^2 \cdot \sin^2(45^\circ)}{2 \cdot 10} \]

Calculating step-by-step:

\[ = \frac{400 \cdot \frac{1}{2}}{20} = \frac{200}{20} = 10 \, \text{m} \]

### Step 4: Determine the radius (R)
At the highest point, the radius \(R\) is equal to the maximum height \(H_{\text{max}}\):

\[ R = H_{\text{max}} = 10 \, \text{m} \]

### Step 5: Calculate the angular velocity (\(\omega\))
The angular velocity \(\omega\) about the point of projection can be calculated using the formula:

\[ \omega = \frac{V}{R} \]

Substituting the values we found:

\[ \omega = \frac{20 \, \text{m/s}}{10 \, \text{m}} = 2 \, \text{rad/s} \]

### Conclusion
The angular velocity of the particle at the highest point of its journey about the point of projection is:

\[ \omega = 2 \, \text{rad/s} \]