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

Options

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

Explanation

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

### Step 1: Understand the Geometry of the Problem
We have an inclined plane at an angle of \( 30^\circ \) to the horizontal. A particle is projected from the foot of this plane at an angle \( \alpha \) above the horizontal.

### Step 2: Set Up the Coordinate System
Let’s set up a coordinate system where:
- The x-axis is along the horizontal.
- The y-axis is vertical.
- The inclined plane makes an angle of \( 30^\circ \) with the horizontal.

### Step 3: Determine the Components of Velocity
When the particle is projected with an initial velocity \( u \) at an angle \( \alpha \):
- The horizontal component of the velocity is \( u \cos \alpha \).
- The vertical component of the velocity is \( u \sin \alpha \).

### Step 4: Calculate the Time of Flight
The time of flight until the particle strikes the inclined plane can be derived from the equations of motion. The vertical motion can be described by:
\[ y = u \sin \alpha \cdot t - \frac{1}{2} g t^2 \]
The equation of the inclined plane is:
\[ y = x \tan(30^\circ) = x \cdot \frac{1}{\sqrt{3}} \]
Substituting \( x = u \cos \alpha \cdot t \) into the equation of the inclined plane gives:
\[ u \sin \alpha \cdot t - \frac{1}{2} g t^2 = (u \cos \alpha \cdot t) \cdot \frac{1}{\sqrt{3}} \]

### Step 5: Find the Condition for Normal Impact
For the particle to strike the inclined plane normally, the vertical component of the velocity at the point of impact must equal the slope of the incline. The vertical component of the velocity at that point is given by:
\[ v_y = u \sin \alpha - g t \]
The slope of the incline is given by \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Therefore, at the point of impact:
\[ v_y = \frac{1}{\sqrt{3}} v_x \]
Where \( v_x = u \cos \alpha \).

### Step 6: Set Up the Equation
From the condition for normal impact, we have:
\[ u \sin \alpha - g t = \frac{1}{\sqrt{3}} (u \cos \alpha) \]

### Step 7: Substitute for Time \( t \)
From the time of flight equation, we can express \( t \) in terms of \( u \) and \( \alpha \):
\[ t = \frac{2u \sin(\alpha - 30^\circ)}{g \cos(30^\circ)} \]

### Step 8: Solve for \( \alpha \)
Substituting \( t \) into the normal impact condition and simplifying leads to:
\[ \tan(\alpha - 30^\circ) = \cot(30^\circ) \cdot \frac{1}{2} \]
This gives:
\[ \tan(\alpha - 30^\circ) = \frac{1}{\sqrt{3}} \cdot \frac{1}{2} \]

### Step 9: Final Calculation
Solving for \( \alpha \):
\[ \alpha - 30^\circ = \tan^{-1}\left(\frac{1}{2\sqrt{3}}\right) \]
Thus:
\[ \alpha = 30^\circ + \tan^{-1}\left(\frac{1}{2\sqrt{3}}\right) \]

### Conclusion
The angle \( \alpha \) at which the particle must be projected to strike the inclined plane normally is:
\[ \alpha = 30^\circ + \tan^{-1}\left(\frac{1}{2\sqrt{3}}\right) \]