If at any instant the velocity of a projectile be u and its direction of motion `theta` to the horizon then it will be moving at right angles to this direction, after time `(u)/(g) cosec theta`

Answer

To solve the problem, we need to show that if at any instant the velocity of a projectile is \( u \) and its direction of motion makes an angle \( \theta \) with the horizontal, then it will be moving at right angles to this direction after a time \( \frac{u}{g} \csc \theta \).

### Step-by-Step Solution:

1. **Initial Velocity Components**:
The initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

2. **Velocity After Time \( t \)**:
After a time \( t \), the horizontal velocity remains constant since there is no horizontal acceleration:
- \( v_x = u \cos \theta \)

The vertical velocity changes due to the acceleration due to gravity \( g \):
- \( v_y = u_y - gt = u \sin \theta - gt \)

3. **Velocity Vector After Time \( t \)**:
The velocity vector after time \( t \) can be expressed as:
\[ \mathbf{v} = v_x \hat{i} + v_y \hat{j} = u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j} \]

4. **Condition for Perpendicularity**:
For the two velocity vectors to be perpendicular, their dot product must be zero:
\[ \mathbf{v_i} \cdot \mathbf{v} = 0 \]
where \( \mathbf{v_i} = u \cos \theta \hat{i} + u \sin \theta \hat{j} \).

5. **Calculating the Dot Product**:
The dot product is calculated as follows:
\[ (u \cos \theta)(u \cos \theta) + (u \sin \theta)(u \sin \theta - gt) = 0 \]
Expanding this gives:
\[ u^2 \cos^2 \theta + u^2 \sin^2 \theta - u g t \sin \theta = 0 \]

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

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

8. **Final Expression**:
We can express \( \sin \theta \) as \( \csc \theta \):
\[ t = \frac{u}{g} \csc \theta \]

This shows that after the time \( \frac{u}{g} \csc \theta \), the projectile will be moving at right angles to its initial direction of motion.