A particle is projected at `t = 0` with velocity u at angle `theta` with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :

Options

A. `(g^(2)sin2theta)/(u^(2))`
B. `(2g^(2)sin theta)/(u^(2))`
C. `(g^(2)sin theta)/(u^(2))`
D. `(g^(2)sin 2theta)/(2u^(2))`

Explanation

To solve the problem, we need to find the ratio of the tangential acceleration and the radius of curvature at the point of projection for a particle projected with an initial velocity \( u \) at an angle \( \theta \) with the horizontal.

### Step-by-Step Solution:

1. **Identify Tangential Acceleration**:
- The only acceleration acting on the particle after projection is due to gravity, which acts vertically downward with a magnitude of \( g \).
- The tangential acceleration \( A_t \) is the component of gravitational acceleration acting along the direction of the velocity of the particle.
- At the point of projection, the angle between the velocity vector and the vertical is \( \theta \). Therefore, the tangential acceleration can be calculated as:
\[ A_t = g \sin \theta \]

2. **Determine the Radius of Curvature**:
- The radius of curvature \( R \) at the point of projection can be calculated using the formula:
\[ R = \frac{V^2}{A_n} \]
- Here, \( V \) is the velocity at the point of projection, which is \( u \), and \( A_n \) is the acceleration perpendicular to the velocity (normal acceleration).
- The normal acceleration is the component of gravitational acceleration acting perpendicular to the velocity, which is given by:
\[ A_n = g \cos \theta \]
- Therefore, substituting these values into the radius of curvature formula gives:
\[ R = \frac{u^2}{g \cos \theta} \]

3. **Calculate the Ratio of Tangential Acceleration to Radius of Curvature**:
- Now we can find the ratio of tangential acceleration to the radius of curvature:
\[ \frac{A_t}{R} = \frac{g \sin \theta}{\frac{u^2}{g \cos \theta}} = \frac{g^2 \sin \theta \cos \theta}{u^2} \]

4. **Simplify the Expression**:
- We can use the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to simplify the expression:
\[ \frac{A_t}{R} = \frac{g^2 \sin 2\theta}{2 u^2} \]

5. **Final Result**:
- Thus, the ratio of the tangential acceleration to the radius of curvature at the point of projection is:
\[ \frac{A_t}{R} = \frac{g^2 \sin 2\theta}{2 u^2} \]

### Conclusion:
The correct answer is:
\[ \frac{g^2 \sin 2\theta}{2 u^2} \]