A particle is projected horizontally from the top of a tower with a velocity `v_(0)`. If v be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directely proportional to

Options

A. `v^(3)`
B. `v^(2)`
C. `v`
D. `1//v`

Explanation

To solve the problem, we need to determine how the radius of curvature (r) of the path of a particle projected horizontally from the top of a tower is related to its velocity (v) at any instant.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle is projected horizontally with an initial velocity \( v_0 \) from the top of a tower. As it falls, it undergoes projectile motion, which consists of horizontal motion with constant velocity and vertical motion under the influence of gravity.

2. **Identifying Forces**:
At any instant, the forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The horizontal component of the velocity \( v_0 \) (which remains constant) and the vertical component of the velocity \( v_y \) which increases due to gravity.

3. **Velocity at Any Instant**:
The total velocity \( v \) of the particle at any instant can be expressed as:
\[ v = \sqrt{v_0^2 + v_y^2} \]
where \( v_y = gt \) is the vertical velocity component after time \( t \).

4. **Centripetal Force and Radius of Curvature**:
The radius of curvature \( r \) of the path at any point can be derived from the centripetal force equation:
\[ F_c = \frac{mv^2}{r} \]
The only force providing centripetal acceleration in this case is the weight of the particle, which is \( mg \). Therefore, we can set:
\[ mg = \frac{mv^2}{r} \]

5. **Solving for Radius of Curvature**:
Rearranging the equation gives us:
\[ r = \frac{v^2}{g} \]
Here, \( g \) is the acceleration due to gravity, which is a constant.

6. **Direct Proportionality**:
From the equation \( r = \frac{v^2}{g} \), we can see that the radius of curvature \( r \) is directly proportional to the square of the velocity \( v \):
\[ r \propto v^2 \]

### Conclusion:
Thus, the radius of curvature of the path of the particle at any instant is directly proportional to \( v^2 \).