A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

Explanation

To solve the problem of finding the time when the velocity vector of a particle projected at an angle \(\theta\) with an initial velocity \(u\) becomes perpendicular to the initial velocity vector, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Velocity Components**:
The initial velocity \(u\) can be broken down into its horizontal and vertical components:
\[ u_x = u \cos \theta \quad \text{(horizontal component)} \]
\[ u_y = u \sin \theta \quad \text{(vertical component)} \]

2. **Write the Velocity Vector as a Function of Time**:
The velocity vector \(\vec{v}(t)\) at any time \(t\) can be expressed as:
\[ \vec{v}(t) = u_x \hat{i} + (u_y - g t) \hat{j} \]
Substituting the components:
\[ \vec{v}(t) = u \cos \theta \hat{i} + (u \sin \theta - g t) \hat{j} \]

3. **Condition for Perpendicular Vectors**:
The condition for two vectors to be perpendicular is that their dot product is zero:
\[ \vec{v}(t) \cdot \vec{u} = 0 \]
Substituting the vectors:
\[ (u \cos \theta \hat{i} + (u \sin \theta - g t) \hat{j}) \cdot (u \cos \theta \hat{i} + u \sin \theta \hat{j}) = 0 \]

4. **Calculate the Dot Product**:
Expanding the dot product:
\[ (u \cos \theta)(u \cos \theta) + (u \sin \theta - g t)(u \sin \theta) = 0 \]
This simplifies to:
\[ u^2 \cos^2 \theta + u \sin \theta (u \sin \theta - g t) = 0 \]
\[ u^2 \cos^2 \theta + u^2 \sin^2 \theta - g u \sin \theta t = 0 \]

5. **Use the Pythagorean Identity**:
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ u^2 - g u \sin \theta t = 0 \]

6. **Solve for Time \(t\)**:
Rearranging gives:
\[ g u \sin \theta t = u^2 \]
\[ t = \frac{u}{g \sin \theta} \]

### Final Result:
Thus, the time when the velocity vector is perpendicular to the initial velocity vector is:
\[ t = \frac{u}{g \sin \theta} \]