A particle is projected with a speed u at angle `theta` with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the radius of curvature of the curve at the point.

Options

A. `(u^2 cos^2 theta)/(2g)`
B. `(sqrt(3) u^2 cos^2 theta)/(2g)`
C. `(u^2 cos^2 theta)/(g)`
D. `((sqrt(3) u^2 cos^2 theta))/(g)`

Explanation

To find the radius of curvature of the projectile's path at its highest point, we can follow these steps:

### Step 1: Understand the motion of the projectile
When a particle is projected with an initial speed \( u \) at an angle \( \theta \) with the horizontal, it follows a parabolic trajectory. The highest point of this trajectory is where the vertical component of the velocity becomes zero.

### Step 2: Identify the velocity components at the highest point
At the highest point of the projectile's motion, the vertical component of the velocity is zero, and only the horizontal component remains. The horizontal component of the initial velocity is given by:
\[ v_x = u \cos \theta \]

### Step 3: Apply the concept of centripetal force
At the highest point, the particle moves in a circular arc. The centripetal force required to keep the particle in circular motion is provided by the gravitational force acting on it. The centripetal force can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is the velocity at the highest point, and \( r \) is the radius of curvature.

### Step 4: Set up the equation for centripetal force
At the highest point, the gravitational force \( mg \) acts as the centripetal force:
\[ \frac{mv^2}{r} = mg \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r} = g \]

### Step 5: Substitute the horizontal velocity into the equation
We know that at the highest point, the horizontal velocity \( v \) is \( u \cos \theta \). Substituting this into the equation gives:
\[ \frac{(u \cos \theta)^2}{r} = g \]

### Step 6: Solve for the radius of curvature \( r \)
Rearranging the equation to solve for \( r \):
\[ r = \frac{(u \cos \theta)^2}{g} \]

### Conclusion
The radius of curvature of the projectile's path at the highest point is given by:
\[ r = \frac{u^2 \cos^2 \theta}{g} \]