A particle of mass m is projecte dwilth speed u at an angle `theta` with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Explanation

To solve the problem of finding the torque of the weight of a particle about the point of projection when the particle is at its highest point, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A particle of mass \( m \) is projected with an initial speed \( u \) at an angle \( \theta \) with the horizontal.
- We need to analyze the situation when the particle reaches its highest point in its projectile motion.

2. **Identify Forces Acting on the Particle**:
- At the highest point, the only force acting on the particle is its weight, which is directed downward. The weight \( W \) is given by:
\[ W = mg \]
- Here, \( g \) is the acceleration due to gravity.

3. **Determine the Position of the Particle at the Highest Point**:
- The horizontal component of the initial velocity is \( u \cos \theta \).
- The vertical component of the initial velocity is \( u \sin \theta \).
- The time taken to reach the highest point \( t \) can be calculated using:
\[ t = \frac{u \sin \theta}{g} \]
- The horizontal distance traveled (which is also the range at the highest point) is:
\[ R = u \cos \theta \cdot t = u \cos \theta \cdot \frac{u \sin \theta}{g} = \frac{u^2 \sin 2\theta}{g} \]

4. **Finding the Torque**:
- The torque \( \tau \) about the point of projection due to the weight of the particle is given by:
\[ \tau = \mathbf{r} \times \mathbf{F} \]
- Here, \( \mathbf{r} \) is the position vector from the point of projection to the particle at the highest point, and \( \mathbf{F} \) is the weight \( mg \).
- At the highest point, the position vector \( \mathbf{r} \) has a horizontal component equal to the range \( R \) and a vertical component of 0.

5. **Calculating the Magnitude of Torque**:
- The angle between \( \mathbf{r} \) (which is horizontal) and the weight \( mg \) (which is vertical) is \( 90^\circ \).
- Thus, the magnitude of the torque is:
\[ \tau = r \cdot mg \cdot \sin(90^\circ) = r \cdot mg \]
- Since \( r = R = \frac{u^2 \sin 2\theta}{g} \), we substitute this into the torque equation:
\[ \tau = \left(\frac{u^2 \sin 2\theta}{g}\right) \cdot mg \]
- Simplifying this gives:
\[ \tau = \frac{u^2 m \sin 2\theta}{g} \]

6. **Final Result**:
- The torque of the weight of the particle about the point of projection when the particle is at the highest point is:
\[ \tau = \frac{u^2 m \sin 2\theta}{g} \]