A particle is projected with a speed u at an angle `theta` with the horizontal . If `R_(1)` is radius of curvature of trajectory at the initial point and `R_(2)` is the radius of curvature at the highest point . Value of `R_2//(R_(1)` is found to be `cos ^(n) theta`. What is the value of n ?

Explanation

To solve the problem, we need to find the ratio of the radii of curvature \( R_2/R_1 \) at the highest point and the initial point of the projectile's trajectory. Let's break this down step by step.

### Step 1: Understanding the Radius of Curvature
The radius of curvature \( R \) at any point in a projectile's trajectory can be determined using the formula:
\[ R = \frac{(1 + (dy/dx)^2)^{3/2}}{|d^2y/dx^2|} \]
where \( dy/dx \) is the slope of the trajectory at that point and \( d^2y/dx^2 \) is the curvature.

### Step 2: Finding \( R_1 \) at the Initial Point
At the initial point of projection (where the angle is \( \theta \)):
- The vertical component of the velocity is \( u \sin \theta \).
- The horizontal component of the velocity is \( u \cos \theta \).

The slope \( dy/dx \) at the initial point can be expressed as:
\[ \frac{dy}{dx} = \tan \theta \]
Thus, the first derivative \( dy/dx \) is \( \tan \theta \).

Now, we need to find \( d^2y/dx^2 \) at this point. The acceleration due to gravity acts downward, which affects the vertical motion. The second derivative \( d^2y/dx^2 \) can be derived from the motion equations, and it will be negative due to the downward acceleration:
\[ \frac{d^2y}{dx^2} = -\frac{g}{u^2 \cos^2 \theta} \]

Now substituting these into the radius of curvature formula for \( R_1 \):
\[ R_1 = \frac{(1 + \tan^2 \theta)^{3/2}}{\left| -\frac{g}{u^2 \cos^2 \theta} \right|} = \frac{(1 + \tan^2 \theta)^{3/2} u^2 \cos^2 \theta}{g} \]
Using \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ R_1 = \frac{(\sec^2 \theta)^{3/2} u^2 \cos^2 \theta}{g} = \frac{u^2 \sec^3 \theta \cos^2 \theta}{g} \]

### Step 3: Finding \( R_2 \) at the Highest Point
At the highest point of the trajectory:
- The vertical component of the velocity is zero, and the horizontal component remains \( u \cos \theta \).

At this point, the slope \( dy/dx \) is zero, which means:
\[ \frac{dy}{dx} = 0 \]
Thus, the first derivative \( dy/dx \) is \( 0 \).

The second derivative \( d^2y/dx^2 \) can be derived similarly, and it will be:
\[ \frac{d^2y}{dx^2} = -\frac{g}{u^2 \cos^2 \theta} \]

Now substituting these into the radius of curvature formula for \( R_2 \):
\[ R_2 = \frac{(1 + 0)^{3/2}}{\left| -\frac{g}{u^2 \cos^2 \theta} \right|} = \frac{u^2 \cos^2 \theta}{g} \]

### Step 4: Finding the Ratio \( R_2/R_1 \)
Now we can find the ratio \( R_2/R_1 \):
\[ \frac{R_2}{R_1} = \frac{\frac{u^2 \cos^2 \theta}{g}}{\frac{u^2 \sec^3 \theta \cos^2 \theta}{g}} = \frac{1}{\sec^3 \theta} = \cos^3 \theta \]

### Conclusion
Thus, we have:
\[ \frac{R_2}{R_1} = \cos^3 \theta \]
This implies that \( n = 3 \).

### Final Answer
The value of \( n \) is \( 3 \).

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