A particle of mass m is projected with an initial velocity U at an angle `theta` to the horizontal. The torque of gravity on projectile at maximum height about the point of projection is

Options

A. `(mU^(2)Sin2theta)/(2)`
B. `mU^(2)Sin2theta`
C. `(mU^(2)Sintheta)/(2)`
D. `(1)/(2)mU^(2)(Sin2theta)^(2)`

Explanation

To solve the problem of finding the torque of gravity on a projectile at its maximum height about the point of projection, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle of mass \( m \) is projected with an initial velocity \( U \) at an angle \( \theta \) to the horizontal.
- We need to find the torque due to gravity when the particle reaches its maximum height.

2. **Identify Forces**:
- The only force acting on the particle at maximum height is the weight of the particle, which is \( F = mg \) (where \( g \) is the acceleration due to gravity).

3. **Determine the Position at Maximum Height**:
- At maximum height, the vertical component of the velocity becomes zero, but the horizontal component remains \( U \cos \theta \).
- The maximum height \( h \) can be calculated using the formula:
\[ h = \frac{U^2 \sin^2 \theta}{2g} \]

4. **Calculate the Horizontal Distance**:
- The range \( R \) of the projectile can be calculated as:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
- Therefore, the horizontal distance to the maximum height is:
\[ \frac{R}{2} = \frac{U^2 \sin 2\theta}{2g} \]

5. **Determine the Torque**:
- The torque \( \tau \) about the point of projection due to the weight of the particle at maximum height is given by:
\[ \tau = r \cdot F \cdot \sin(\phi) \]
- Here, \( r \) is the horizontal distance to the point of projection at maximum height (which is \( \frac{R}{2} \)), \( F = mg \), and \( \phi \) is the angle between the position vector and the force vector.
- At maximum height, the angle \( \phi \) is \( 90^\circ \) (since the weight acts downward and the position vector is horizontal), thus \( \sin(90^\circ) = 1 \).

6. **Substituting Values**:
- Now substituting the values into the torque equation:
\[ \tau = \left(\frac{U^2 \sin 2\theta}{2g}\right) \cdot mg \cdot 1 \]
- Simplifying gives:
\[ \tau = \frac{m U^2 \sin 2\theta}{2} \]

### Final Answer:
The torque of gravity on the projectile at maximum height about the point of projection is:
\[ \tau = \frac{m U^2 \sin 2\theta}{2} \]