A particle is projected with a speed of `10 m//s` at an angle `37^(@)` from horizontal, angular speed to particle with respect to point of projection at `t = 0.5 sec`. Is `((16X)/(61))` rad/sec then calculate `X` :

Explanation

To solve the problem step by step, we will calculate the angular speed of the particle projected at a speed of 10 m/s at an angle of 37 degrees with respect to the point of projection at t = 0.5 seconds.

### Step 1: Resolve the initial velocity into components
The initial velocity \( u = 10 \, \text{m/s} \) is projected at an angle of \( 37^\circ \).

- The horizontal component of the velocity \( u_x \) is given by:
\[ u_x = u \cos(37^\circ) = 10 \times \frac{4}{5} = 8 \, \text{m/s} \]

- The vertical component of the velocity \( u_y \) is given by:
\[ u_y = u \sin(37^\circ) = 10 \times \frac{3}{5} = 6 \, \text{m/s} \]

### Step 2: Calculate the position of the particle at \( t = 0.5 \, \text{s} \)
- For the horizontal position \( x \):
\[ x = u_x \cdot t = 8 \times 0.5 = 4 \, \text{m} \]

- For the vertical position \( y \):
Using the equation of motion:
\[ y = u_y \cdot t + \frac{1}{2} a_y t^2 \]
where \( a_y = -10 \, \text{m/s}^2 \) (acceleration due to gravity),
\[ y = 6 \times 0.5 + \frac{1}{2} \times (-10) \times (0.5)^2 \]
\[ y = 3 - 1.25 = 1.75 \, \text{m} \]

### Step 3: Determine the position vector \( \mathbf{R} \)
The position vector \( \mathbf{R} \) at \( t = 0.5 \, \text{s} \) is:
\[ \mathbf{R} = 4 \hat{i} + 1.75 \hat{j} \, \text{m} \]

### Step 4: Calculate the velocity vector \( \mathbf{V} \) at \( t = 0.5 \, \text{s} \)
- The horizontal velocity remains constant:
\[ V_x = u_x = 8 \, \text{m/s} \]

- The vertical velocity \( V_y \) at \( t = 0.5 \, \text{s} \):
\[ V_y = u_y + a_y \cdot t = 6 - 10 \times 0.5 = 1 \, \text{m/s} \]

Thus, the velocity vector \( \mathbf{V} \) is:
\[ \mathbf{V} = 8 \hat{i} + 1 \hat{j} \, \text{m/s} \]

### Step 5: Calculate the angular velocity \( \omega \)
The angular velocity \( \omega \) can be found using the formula:
\[ \omega = \frac{|\mathbf{V} \times \mathbf{R}|}{|\mathbf{R}|^2} \]

- First, calculate \( \mathbf{V} \times \mathbf{R} \):
\[ \mathbf{V} \times \mathbf{R} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 1 & 0 \\ 4 & 1.75 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ = (8 \cdot 1.75 - 1 \cdot 4) \hat{k} = (14 - 4) \hat{k} = 10 \hat{k} \]
Thus, \( |\mathbf{V} \times \mathbf{R}| = 10 \).

- Now, calculate \( |\mathbf{R}|^2 \):
\[ |\mathbf{R}|^2 = 4^2 + 1.75^2 = 16 + 3.0625 = 19.0625 \]

### Step 6: Substitute into the angular velocity formula
\[ \omega = \frac{10}{19.0625} \]

### Step 7: Set the equation equal to the given angular speed
We are given that:
\[ \omega = \frac{16X}{61} \]
Setting the two expressions for \( \omega \) equal:
\[ \frac{10}{19.0625} = \frac{16X}{61} \]

### Step 8: Solve for \( X \)
Cross-multiplying gives:
\[ 10 \cdot 61 = 16X \cdot 19.0625 \]
\[ 610 = 16X \cdot 19.0625 \]
\[ X = \frac{610}{16 \cdot 19.0625} \]
Calculating \( 16 \cdot 19.0625 = 305 \):
\[ X = \frac{610}{305} = 2 \]

### Final Answer
Thus, the value of \( X \) is:
\[ \boxed{2} \]