A stone is projected at a speed `v_(0)` and it makes an angle `theta` with the horizontal. Find the angular velocity of the projectile at the time when it reaches the same level with respect to the point of projection at the time it touches the ground.

Answer

To find the angular velocity of a projectile at the moment it reaches the same level as the point of projection, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- A stone is projected with an initial speed \( v_0 \) at an angle \( \theta \) with the horizontal. We need to find the angular velocity \( \omega \) when the stone returns to the same height from which it was projected.

2. **Identify Components of Velocity**:
- The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component: \( v_{0x} = v_0 \cos \theta \)
- Vertical component: \( v_{0y} = v_0 \sin \theta \)

3. **Determine the Velocity at the Same Level**:
- When the stone returns to the same level, the horizontal component of the velocity remains unchanged: \( v_{x} = v_0 \cos \theta \).
- The vertical component of the velocity just before hitting the ground will be \( v_{y} = -v_0 \sin \theta \) (the negative sign indicates that it is directed downwards).

4. **Calculate the Perpendicular Velocity**:
- The perpendicular component of the velocity with respect to the point of projection at the moment of impact is the vertical component:
\[ v_{\perp} = v_{0} \sin \theta \]

5. **Determine the Distance \( r \)**:
- The distance \( r \) from the point of projection to the point where the stone hits the ground is the range of the projectile, given by:
\[ R = \frac{v_0^2 \sin 2\theta}{g} \]

6. **Calculate Angular Velocity \( \omega \)**:
- The formula for angular velocity \( \omega \) is given by:
\[ \omega = \frac{v_{\perp}}{r} \]
- Substituting the values we have:
\[ \omega = \frac{v_0 \sin \theta}{\frac{v_0^2 \sin 2\theta}{g}} \]
- Simplifying this expression:
\[ \omega = \frac{v_0 \sin \theta \cdot g}{v_0^2 \sin 2\theta} \]
- Since \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite:
\[ \omega = \frac{g}{2 v_0 \cos \theta} \]

7. **Final Result**:
- Thus, the angular velocity \( \omega \) of the projectile at the moment it touches the ground is:
\[ \omega = \frac{g}{2 v_0 \cos \theta} \]