Two particles A and B are projected simultaneously from a fixed point of the ground. Particle A is projected on a smooth horizontal surface with speed v, while particle B is projected in air with speed v, while particle B is projected in air with speed `(2v)/(sqrt(3))`. If particle B hits the particle A, the angle of projection of B with the vertical is

Options

A. `30^(@)`
B. `60^(@)`
C. Both (a) and (b)
D. `45^(@)`

Explanation

To solve the problem, we need to determine the angle of projection of particle B with respect to the vertical when it hits particle A. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the motion of both particles
- Particle A is projected horizontally with speed \( v \).
- Particle B is projected at an angle \( \theta \) with speed \( \frac{2v}{\sqrt{3}} \).

### Step 2: Set up the equations for horizontal displacement
- The horizontal displacement of particle A after time \( t \) is given by:
\[ x_A = vt \]
- The horizontal displacement of particle B is given by:
\[ x_B = \left(\frac{2v}{\sqrt{3}} \cos \theta\right) t \]

### Step 3: Set the horizontal displacements equal to each other
Since both particles hit each other at the same horizontal position, we can equate their displacements:
\[ vt = \left(\frac{2v}{\sqrt{3}} \cos \theta\right) t \]

### Step 4: Simplify the equation
We can cancel \( t \) from both sides (assuming \( t \neq 0 \)):
\[ v = \frac{2v}{\sqrt{3}} \cos \theta \]
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 1 = \frac{2}{\sqrt{3}} \cos \theta \]

### Step 5: Solve for \( \cos \theta \)
Rearranging gives:
\[ \cos \theta = \frac{\sqrt{3}}{2} \]

### Step 6: Find the angle \( \theta \)
The angle \( \theta \) for which \( \cos \theta = \frac{\sqrt{3}}{2} \) is:
\[ \theta = 30^\circ \]

### Step 7: Find the angle with respect to the vertical
Since the angle of projection is given with respect to the horizontal, to find the angle with respect to the vertical, we subtract \( \theta \) from \( 90^\circ \):
\[ \text{Angle with vertical} = 90^\circ - 30^\circ = 60^\circ \]

### Final Answer
The angle of projection of particle B with the vertical is \( 60^\circ \).

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