A particle is projected with a velocity u making an angle `theta` with the horizontal. At any instant its velocity becomes v which is perpendicular to the initial velocity u. Then v is

Options

A. `u cos theta`
B. `u tan theta`
C. `u sec theta`
D. `ucot theta`

Explanation

To solve the problem, we need to analyze the motion of a particle projected at an angle with an initial velocity and determine the conditions when its velocity becomes perpendicular to the initial velocity.

### Step-by-Step Solution:

1. **Understanding the Initial Velocity**:
The particle is projected with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. We can resolve this initial velocity into its horizontal and vertical components:
- Horizontal component: \( u_x = u \cos(\theta) \)
- Vertical component: \( u_y = u \sin(\theta) \)

2. **Velocity at Any Instant**:
As the particle moves under the influence of gravity, its vertical component of velocity changes due to gravitational acceleration. The vertical component of velocity \( v_y \) at any time \( t \) can be expressed as:
\[ v_y = u_y - g t = u \sin(\theta) - g t \]
The horizontal component of velocity remains constant:
\[ v_x = u_x = u \cos(\theta) \]

3. **Condition for Perpendicularity**:
The velocity \( v \) becomes perpendicular to the initial velocity \( u \) when the dot product of the two velocity vectors is zero. This can be expressed mathematically as:
\[ u_x \cdot v_x + u_y \cdot v_y = 0 \]
Substituting the components:
\[ (u \cos(\theta))(u \cos(\theta)) + (u \sin(\theta))(u \sin(\theta) - g t) = 0 \]

4. **Simplifying the Equation**:
Expanding the equation gives:
\[ u^2 \cos^2(\theta) + u \sin(\theta)(u \sin(\theta) - g t) = 0 \]
This simplifies to:
\[ u^2 \cos^2(\theta) + u^2 \sin^2(\theta) - u g t \sin(\theta) = 0 \]
Using the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ u^2 - u g t \sin(\theta) = 0 \]

5. **Finding Time \( t \)**:
Rearranging gives:
\[ u g t \sin(\theta) = u^2 \]
Thus,
\[ t = \frac{u}{g \sin(\theta)} \]

6. **Calculating the Velocity \( v \)**:
Now we can find the magnitude of the velocity \( v \) when it is perpendicular to \( u \). The magnitude of the velocity \( v \) can be calculated using:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the components:
\[ v_x = u \cos(\theta) \quad \text{and} \quad v_y = u \sin(\theta) - g t \]
Substituting \( t \):
\[ v_y = u \sin(\theta) - g \left(\frac{u}{g \sin(\theta)}\right) = u \sin(\theta) - u \frac{1}{\sin(\theta)} = u \left(\sin(\theta) - \frac{1}{\sin(\theta)}\right) \]

7. **Final Expression for \( v \)**:
Therefore, the magnitude of the velocity \( v \) when it is perpendicular to the initial velocity \( u \) is:
\[ v = \sqrt{(u \cos(\theta))^2 + \left(u \left(\sin(\theta) - \frac{1}{\sin(\theta)}\right)\right)^2} \]