A body is projected at `60^(@)` with the horizontal with the velocity of `10sqrt(3)ms^(-1)`. Find the velocity of the projectile when it moves perpendicular to its initial direction.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Identify the given values
- The angle of projection, \( \theta = 60^\circ \)
- The initial velocity, \( u = 10\sqrt{3} \, \text{m/s} \)

### Step 2: Understand the problem
We need to find the velocity \( v \) of the projectile when it moves perpendicular to its initial direction. When the projectile moves perpendicular to its initial direction, the angle between the initial velocity vector and the final velocity vector will be \( 90^\circ \).

### Step 3: Set up the relationship between the components of velocity
The horizontal component of the initial velocity \( u \) is given by:
\[ u_x = u \cos \theta \]
The vertical component of the initial velocity \( u \) is given by:
\[ u_y = u \sin \theta \]

When the projectile moves perpendicular to the initial direction, we can denote the final velocity \( v \) at that point. The horizontal component of \( v \) will be:
\[ v_x = v \cos(90^\circ - \theta) = v \sin \theta \]

### Step 4: Use the conservation of horizontal velocity
Since there is no horizontal acceleration (assuming no air resistance), the horizontal component of the initial velocity must equal the horizontal component of the final velocity:
\[ u \cos \theta = v \sin \theta \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ 10\sqrt{3} \cos(60^\circ) = v \sin(60^\circ) \]

### Step 6: Calculate the trigonometric values
We know:
\[ \cos(60^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Substituting these values into the equation:
\[ 10\sqrt{3} \cdot \frac{1}{2} = v \cdot \frac{\sqrt{3}}{2} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 5\sqrt{3} = v \cdot \frac{\sqrt{3}}{2} \]

### Step 8: Solve for \( v \)
To isolate \( v \), multiply both sides by \( 2/\sqrt{3} \):
\[ v = 5\sqrt{3} \cdot \frac{2}{\sqrt{3}} = 10 \, \text{m/s} \]

### Conclusion
The velocity of the projectile when it moves perpendicular to its initial direction is:
\[ \boxed{10 \, \text{m/s}} \]