A particle is projected with initial speed u at an angle `theta` above the horizontal . Let A be the point of projection , B the point where velocity makes an angle `theta //2` above the horizontal and C the highest point of the trajectory .
Radius of curvature of the trajectory at point C is

Options

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

Explanation

To find the radius of curvature of the trajectory at point C (the highest point of the projectile motion), we can follow these steps:

### Step 1: Understand the motion of the projectile
The projectile is launched with an initial speed \( u \) at an angle \( \theta \) above the horizontal. The trajectory of the projectile is parabolic.

### Step 2: Determine the velocity at point C
At the highest point C, the vertical component of the velocity becomes zero. The horizontal component of the velocity remains unchanged throughout the motion. The horizontal component \( u_x \) is given by:
\[ u_x = u \cos(\theta) \]
Thus, at point C, the velocity \( v_C \) is:
\[ v_C = u \cos(\theta) \]

### Step 3: Calculate the acceleration at point C
The only acceleration acting on the projectile is due to gravity, which acts downwards with an acceleration \( g \). At point C, the acceleration is:
\[ a = g \]

### Step 4: Use the formula for radius of curvature
The radius of curvature \( R \) at any point in projectile motion can be calculated using the formula:
\[ R = \frac{(v^2)}{a} \]
where \( v \) is the velocity at that point and \( a \) is the acceleration.

### Step 5: Substitute the values into the formula
At point C, we substitute \( v = u \cos(\theta) \) and \( a = g \):
\[ R = \frac{(u \cos(\theta))^2}{g} \]

### Step 6: Simplify the expression
This simplifies to:
\[ R = \frac{u^2 \cos^2(\theta)}{g} \]

### Final Answer
The radius of curvature of the trajectory at point C is:
\[ R = \frac{u^2 \cos^2(\theta)}{g} \]

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