A particle is projected with a certain velocity at an angle `prop` above the horizontal from the foot of an inclined plane of inclination `30^@`. If the particle strikes the plane normally, then `prop` is equal to.

Options

A. `30^@ + tan^-1 (sqrt(3))/(2))`
B. `45^@`
C. `60^@`
D. `30^@ + tan^-1(2 sqrt(3))`

Explanation

To solve the problem, we need to find the angle of projection \( \alpha \) such that a particle projected at this angle strikes an inclined plane of inclination \( 30^\circ \) normally. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
We have a particle projected with an initial velocity \( u \) at an angle \( \alpha \) above the horizontal. The inclined plane makes an angle of \( 30^\circ \) with the horizontal. We need to find the angle \( \alpha \) such that the particle strikes the plane normally.

### Step 2: Define the Angles
When the particle strikes the inclined plane normally, the angle of the particle's velocity vector with respect to the plane is \( 90^\circ \). The angle of the velocity vector with respect to the horizontal is \( \alpha \), and the angle of the inclined plane with respect to the horizontal is \( 30^\circ \). Therefore, the angle of the velocity vector with respect to the inclined plane is:
\[ \theta = \alpha - 30^\circ \]

### Step 3: Components of Velocity
The horizontal and vertical components of the initial velocity \( u \) can be expressed as:
- \( u_x = u \cos(\alpha) \) (horizontal component)
- \( u_y = u \sin(\alpha) \) (vertical component)

### Step 4: Acceleration Components
The acceleration due to gravity \( g \) acts downwards. We need to resolve \( g \) into components parallel and perpendicular to the inclined plane:
- The component of \( g \) parallel to the incline is \( g \sin(30^\circ) = \frac{g}{2} \)
- The component of \( g \) perpendicular to the incline is \( g \cos(30^\circ) = g \frac{\sqrt{3}}{2} \)

### Step 5: Condition for Normal Impact
For the particle to strike the plane normally, the horizontal component of the velocity at the moment of impact must be zero. This occurs when the vertical component of the velocity equals the component of gravity acting perpendicular to the incline.

Setting up the equations, we have:
\[ u \sin(\alpha - 30^\circ) = g \cos(30^\circ) t \]
And for the horizontal component:
\[ u \cos(\alpha - 30^\circ) = g \sin(30^\circ) t \]

### Step 6: Time of Flight
From the vertical motion equation:
\[ t = \frac{u \sin(\alpha - 30^\circ)}{g \cos(30^\circ)} \]
From the horizontal motion equation:
\[ t = \frac{u \cos(\alpha - 30^\circ)}{g \sin(30^\circ)} \]

### Step 7: Equating the Times
Since both expressions for \( t \) are equal, we can set them equal to each other:
\[ \frac{u \sin(\alpha - 30^\circ)}{g \cos(30^\circ)} = \frac{u \cos(\alpha - 30^\circ)}{g \sin(30^\circ)} \]

### Step 8: Simplifying the Equation
Cancelling \( u \) and \( g \) (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \frac{\sin(\alpha - 30^\circ)}{\cos(\alpha - 30^\circ)} = \frac{\cos(30^\circ)}{\sin(30^\circ)} \]
This simplifies to:
\[ \tan(\alpha - 30^\circ) = \frac{\sqrt{3}}{1} = \sqrt{3} \]

### Step 9: Solving for \( \alpha \)
Taking the inverse tangent:
\[ \alpha - 30^\circ = 60^\circ \quad \Rightarrow \quad \alpha = 90^\circ \]

### Conclusion
Thus, the angle of projection \( \alpha \) is:
\[ \alpha = 90^\circ \]