A particle is projected from ground at an angle `theta` with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is -

Options

A. `(1)/(sin^(2) theta cos theta)`
B. `cos^(2) theta`
C. `(1)/(sin^(3) theta)`
D. `(1)/(cos^(3) theta)`

Answer

To solve the problem of finding the ratio of the radius of curvature of a projectile's trajectory at the point of projection to that at its maximum height, we can follow these steps:

### Step 1: Understand the Motion of the Projectile
A particle is projected from the ground at an angle \( \theta \) with an initial speed \( u \). The motion can be broken down into horizontal and vertical components.

### Step 2: Determine the Radius of Curvature at the Point of Projection
At the point of projection, the radius of curvature \( R_1 \) can be calculated using the formula:
\[ R_1 = \frac{u^2}{a \cos \theta} \]
Here, \( a \) is the acceleration due to gravity \( g \), and \( \cos \theta \) accounts for the angle of projection.

### Step 3: Determine the Radius of Curvature at Maximum Height
At maximum height, the vertical component of velocity becomes zero, and only the horizontal component \( u \cos \theta \) remains. The radius of curvature \( R_2 \) at maximum height is given by:
\[ R_2 = \frac{(u \cos \theta)^2}{g} \]
This simplifies to:
\[ R_2 = \frac{u^2 \cos^2 \theta}{g} \]

### Step 4: Calculate the Ratio of the Two Radii of Curvature
Now, we can find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{\frac{u^2}{g \cos \theta}}{\frac{u^2 \cos^2 \theta}{g}} = \frac{u^2}{g \cos \theta} \cdot \frac{g}{u^2 \cos^2 \theta} \]
This simplifies to:
\[ \frac{R_1}{R_2} = \frac{1}{\cos^3 \theta} \]

### Final Answer
Thus, the ratio of the radius of curvature at the point of projection to the radius of curvature at maximum height is:
\[ \frac{R_1}{R_2} = \frac{1}{\cos^3 \theta} \]