A particle is projected with speed `u` at angle `theta` to the horizontal. Find the radius of curvature at highest point of its trajectory

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 at the highest point of the trajectory of a particle projected with speed `u` at an angle `theta` to the horizontal, we can follow these steps:

### Step 1: Understand the Components of Velocity
When a particle is projected at an angle `theta`, its initial velocity `u` can be broken down into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

At the highest point of the trajectory, the vertical component of the velocity becomes zero (\( v_y = 0 \)), while the horizontal component remains the same (\( v_x = u \cos \theta \)).

### Step 2: Identify the Forces Acting on the Particle
At the highest point, the only force acting on the particle is the weight of the particle, which acts downwards. The gravitational force \( F_g \) is given by:
\[ F_g = mg \]
where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity.

### Step 3: Relate the Forces to Centripetal Force
At the highest point, the gravitational force provides the necessary centripetal force for the circular motion of the particle. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the horizontal velocity at the highest point and \( r \) is the radius of curvature.

### Step 4: Set Up the Equation
Since the gravitational force equals the centripetal force at the highest point, we can set up the equation:
\[ mg = \frac{mv^2}{r} \]

### Step 5: Solve for Radius of Curvature
We can cancel the mass \( m \) from both sides:
\[ g = \frac{v^2}{r} \]
Rearranging gives:
\[ r = \frac{v^2}{g} \]

### Step 6: Substitute the Velocity
At the highest point, the horizontal velocity \( v \) is:
\[ v = u \cos \theta \]
Substituting this into the equation for \( r \):
\[ r = \frac{(u \cos \theta)^2}{g} \]

### Step 7: Simplify the Expression
This simplifies to:
\[ r = \frac{u^2 \cos^2 \theta}{g} \]

### Final Answer
Thus, the radius of curvature at the highest point of the trajectory is:
\[ r = \frac{u^2 \cos^2 \theta}{g} \]

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