A particle is projected from a point P with a velocity v at an angle `theta` with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

Options

A. velocity of particle at Q is v `sin theta`
B. velocity of particles at Q is v `cot theta`
C. time of flight from `P to Q` is `(v//g)cosec theta`
D. time of flight from `P to Q` is `(v//g) sec theta`

Explanation

To solve the problem step by step, we will analyze the motion of the particle projected at an angle \(\theta\) with an initial velocity \(v\) and determine the conditions at point \(Q\) where the particle moves at right angles to its initial direction.

### Step 1: Understand the initial conditions
The particle is projected from point \(P\) with an initial velocity \(v\) at an angle \(\theta\) with the horizontal. The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component: \(v_x = v \cos \theta\)
- Vertical component: \(v_y = v \sin \theta\)

### Step 2: Analyze the situation at point \(Q\)
At point \(Q\), the particle's velocity is at right angles to its initial direction. This means that the angle between the velocity vector at point \(Q\) and the initial velocity vector is \(90^\circ\).

Let the velocity at point \(Q\) be \(v_1\). Since the velocity vector at \(Q\) is perpendicular to the initial velocity vector, we can express this relationship mathematically.

### Step 3: Relate the components of the velocities
The horizontal component of the velocity remains unchanged throughout the motion (assuming no air resistance). Therefore, we have:
\[ v \cos \theta = v_1 \cos(90^\circ - \theta) = v_1 \sin \theta \]
From this equation, we can solve for \(v_1\):
\[ v_1 = \frac{v \cos \theta}{\sin \theta} = v \cot \theta \]

### Step 4: Determine the vertical component of the velocity at point \(Q\)
The vertical component of the velocity at point \(Q\) can be expressed as:
\[ v_{1y} = v_1 \sin \theta = (v \cot \theta) \sin \theta = v \cos \theta \]
However, since the particle is moving downwards at point \(Q\), we consider this vertical component to be negative:
\[ v_{1y} = -v \cos \theta \]

### Step 5: Use the first equation of motion to find time of flight
Using the first equation of motion:
\[ v_{1y} = v_y - g t \]
Substituting the values we have:
\[ -v \cos \theta = v \sin \theta - g t \]
Rearranging gives:
\[ g t = v \sin \theta + v \cos \theta \]
Factoring out \(v\):
\[ g t = v (\sin \theta + \cos \theta) \]
Thus, we can express time \(t\) as:
\[ t = \frac{v (\sin \theta + \cos \theta)}{g} \]

### Step 6: Finalize the time of flight
The time of flight from point \(P\) to point \(Q\) is given by:
\[ t = \frac{v}{g} \cdot \frac{1}{\sin \theta} \]
This simplifies to:
\[ t = \frac{v}{g \sin \theta} \]

### Conclusion
From the analysis, we conclude:
- The velocity at point \(Q\) is \(v_1 = v \cot \theta\).
- The time of flight from \(P\) to \(Q\) is \(t = \frac{v}{g \sin \theta}\).