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 projectile is perpendicular to its initial velocity vector, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Velocity**:
The initial velocity \( \vec{u} \) of the particle can be resolved into its horizontal and vertical components:
\[ \vec{u} = u \cos \theta \, \hat{i} + u \sin \theta \, \hat{j} \]

2. **Acceleration Due to Gravity**:
The only acceleration acting on the particle is due to gravity, which acts downward:
\[ \vec{a} = -g \, \hat{j} \]

3. **Finding the Final Velocity**:
The velocity \( \vec{v} \) of the particle at time \( t \) can be expressed as:
\[ \vec{v} = \vec{u} + \vec{a} t = (u \cos \theta \, \hat{i} + u \sin \theta \, \hat{j}) + (-g \, \hat{j}) t \]
Simplifying this gives:
\[ \vec{v} = u \cos \theta \, \hat{i} + (u \sin \theta - g t) \, \hat{j} \]

4. **Condition for Perpendicular Vectors**:
The vectors \( \vec{u} \) and \( \vec{v} \) are perpendicular when their dot product is zero:
\[ \vec{u} \cdot \vec{v} = 0 \]
This can be expressed as:
\[ (u \cos \theta) (u \cos \theta) + (u \sin \theta)(u \sin \theta - g t) = 0 \]

5. **Expanding the Dot Product**:
Expanding the dot product gives:
\[ u^2 \cos^2 \theta + u^2 \sin^2 \theta - u g t \sin \theta = 0 \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ u^2 - u g t \sin \theta = 0 \]

6. **Solving for Time \( t \)**:
Rearranging the equation to solve for \( t \):
\[ u g t \sin \theta = u^2 \]
\[ t = \frac{u}{g \sin \theta} \]

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