A particle of mass `m` is projected with velocity `u` at an angle `alpha` with horizontal. During the period when the particle descends from highest point to the position where
its velocity vector makes an angle `(alpha)/(2)` with horizontal. Work done by gravity force is

Options

A. `1/2 m u ^(2) tan^(2)(theta//2)`
B. `1/2 m u^(2)tan^(2)theta`
C. `1/2 m u^(2) cos^(2)theta tan^(2)(theta//2)`
D. `1/2 m u^(2)cos^(2)(theta//2)sin^(2)theta`

Explanation

To find the work done by the gravitational force on a particle of mass \( m \) projected with an initial velocity \( u \) at an angle \( \alpha \) with the horizontal, we will analyze the motion of the particle from its highest point to the position where its velocity vector makes an angle \( \frac{\alpha}{2} \) with the horizontal.

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions**:
- The particle is projected with an initial velocity \( u \) at an angle \( \alpha \).
- At the highest point, the vertical component of the velocity is zero, and the horizontal component is \( u \cos \alpha \).
- We need to find the work done by gravity when the particle descends to a position where its velocity makes an angle \( \frac{\alpha}{2} \) with the horizontal.

2. **Determine the Final Velocity**:
- At the position where the velocity makes an angle \( \frac{\alpha}{2} \), the horizontal component of the velocity remains \( u \cos \alpha \).
- Let the final velocity be \( V \). The horizontal component of the velocity can be expressed as:
\[ V \cos\left(\frac{\alpha}{2}\right) = u \cos \alpha \]
- Therefore, we can express \( V \) as:
\[ V = \frac{u \cos \alpha}{\cos\left(\frac{\alpha}{2}\right)} \]

3. **Calculate Initial and Final Kinetic Energy**:
- The initial kinetic energy \( K_i \) at the highest point is:
\[ K_i = \frac{1}{2} m (u \cos \alpha)^2 = \frac{1}{2} m u^2 \cos^2 \alpha \]
- The final kinetic energy \( K_f \) when the particle makes an angle \( \frac{\alpha}{2} \) is:
\[ K_f = \frac{1}{2} m V^2 = \frac{1}{2} m \left(\frac{u \cos \alpha}{\cos\left(\frac{\alpha}{2}\right)}\right)^2 \]
- Simplifying \( K_f \):
\[ K_f = \frac{1}{2} m \frac{u^2 \cos^2 \alpha}{\cos^2\left(\frac{\alpha}{2}\right)} \]

4. **Calculate the Change in Kinetic Energy**:
- The work done by gravity \( W \) is equal to the change in kinetic energy:
\[ W = K_f - K_i \]
- Substituting the expressions for \( K_f \) and \( K_i \):
\[ W = \frac{1}{2} m \frac{u^2 \cos^2 \alpha}{\cos^2\left(\frac{\alpha}{2}\right)} - \frac{1}{2} m u^2 \cos^2 \alpha \]
- Factoring out \( \frac{1}{2} m u^2 \cos^2 \alpha \):
\[ W = \frac{1}{2} m u^2 \cos^2 \alpha \left(\frac{1}{\cos^2\left(\frac{\alpha}{2}\right)} - 1\right) \]

5. **Simplifying the Expression**:
- Using the identity \( 1 - \cos^2\theta = \sin^2\theta \):
\[ W = \frac{1}{2} m u^2 \cos^2 \alpha \cdot \left(\frac{\sin^2\left(\frac{\alpha}{2}\right)}{\cos^2\left(\frac{\alpha}{2}\right)}\right) \]
- This can be rewritten as:
\[ W = \frac{1}{2} m u^2 \cos^2 \alpha \tan^2\left(\frac{\alpha}{2}\right) \]

### Final Answer:
The work done by the gravitational force is:
\[ W = \frac{1}{2} m u^2 \cos^2 \alpha \tan^2\left(\frac{\alpha}{2}\right) \]