A particle is projected with a speed `u` at an angle `theta` to the horizontal. Find the radius of curvature.
At the point where the particle is at a highest half of the maximum height `H` attained by it.

Options

A. `(2 u^2(1 + cos^2 theta)^(3//2))/(g 2 sqrt(2) cos theta)`
B. `(u^2(1 + cos^2 theta)^(3//2))/(g 2 sqrt(2) cos theta)`
C. `(u^2(1 - sin^2 theta)^(3//2))/(g 2 sqrt(2) cos theta)`
D. `(u^2(1 - tan^2 theta)^(3//2))/(g 2 sqrt(2) cos theta)`

Explanation

To find the radius of curvature of a projectile at half of its maximum height, we can follow these steps:

### Step 1: Determine the maximum height (H) of the projectile
The maximum height \( H \) attained by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the height at which we need to find the radius of curvature
We need to find the radius of curvature at half of the maximum height:
\[ h = \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} \]

### Step 3: Find the velocity (V) at height \( h \)
Using the third equation of motion for the vertical component of velocity, we have:
\[ V_y^2 = u_y^2 - 2gh \]
where \( u_y = u \sin \theta \) is the initial vertical component of velocity. Thus,
\[ V_y^2 = (u \sin \theta)^2 - 2g \left(\frac{u^2 \sin^2 \theta}{4g}\right) \]
Simplifying this gives:
\[ V_y^2 = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2} \]
So,
\[ V_y = \sqrt{\frac{u^2 \sin^2 \theta}{2}} = \frac{u \sin \theta}{\sqrt{2}} \]

The horizontal component of velocity remains constant:
\[ V_x = u \cos \theta \]

### Step 4: Find the resultant velocity (V) at height \( h \)
The resultant velocity \( V \) can be found using Pythagoras' theorem:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(u \cos \theta)^2 + \left(\frac{u \sin \theta}{\sqrt{2}}\right)^2} \]
This simplifies to:
\[ V = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}} = u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 5: Find the centripetal acceleration (A_c)
The centripetal acceleration \( A_c \) at height \( h \) is given by:
\[ A_c = g \cos \alpha \]
where \( \alpha \) is the angle of the velocity vector with the horizontal. We can find \( \cos \alpha \) using:
\[ \cos \alpha = \frac{V_x}{V} = \frac{u \cos \theta}{u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}} = \frac{\cos \theta}{\sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}} \]
Thus,
\[ A_c = g \frac{\cos \theta}{\sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}} \]

### Step 6: Calculate the radius of curvature (R)
The radius of curvature \( R \) is given by the formula:
\[ R = \frac{V^2}{A_c} \]
Substituting the values we found:
\[ R = \frac{u^2 \left(\cos^2 \theta + \frac{\sin^2 \theta}{2}\right)}{g \frac{\cos \theta}{\sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}}} \]
This simplifies to:
\[ R = \frac{u^2 \left(\cos^2 \theta + \frac{\sin^2 \theta}{2}\right) \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}}{g \cos \theta} \]

### Final Result
Thus, the radius of curvature at half of the maximum height is:
\[ R = \frac{u^2 \left(\cos^2 \theta + \frac{\sin^2 \theta}{2}\right) \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}}}{g \cos \theta} \]