A particle is projected upwards with a velocity of `100 m//sec` at an angle of `60^(@)` with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking `g = 10m//sec^(2)`.

Explanation

To solve the problem of finding the time when a particle projected upwards at a velocity of 100 m/s at an angle of 60 degrees with the vertical moves perpendicular to its initial direction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity:**
The initial velocity \( u = 100 \, \text{m/s} \) is given at an angle of \( 60^\circ \) with the vertical. We need to resolve this velocity into its vertical and horizontal components.

- Vertical component \( u_y = u \cos(60^\circ) \)
- Horizontal component \( u_x = u \sin(60^\circ) \)

Using the values:
\[ u_y = 100 \cos(60^\circ) = 100 \times \frac{1}{2} = 50 \, \text{m/s} \]
\[ u_x = 100 \sin(60^\circ) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{m/s} \]

2. **Determine the Velocity Components at Time \( t \):**
The horizontal component of velocity remains constant since there is no horizontal acceleration:
\[ v_x = u_x = 50\sqrt{3} \, \text{m/s} \]

The vertical component of velocity changes due to gravity:
\[ v_y = u_y - g t = 50 - 10t \, \text{m/s} \]

3. **Condition for Perpendicular Motion:**
The particle will move perpendicular to its initial direction when the dot product of the initial velocity vector and the current velocity vector is zero. This means:
\[ u_x v_x + u_y v_y = 0 \]

Substituting the values:
\[ (50\sqrt{3})(50\sqrt{3}) + (50)(50 - 10t) = 0 \]

4. **Simplify the Equation:**
\[ 2500 \cdot 3 + 2500 - 500t = 0 \]
\[ 7500 + 2500 - 500t = 0 \]
\[ 10000 - 500t = 0 \]

5. **Solve for Time \( t \):**
\[ 500t = 10000 \]
\[ t = \frac{10000}{500} = 20 \, \text{seconds} \]

### Final Answer:
The time when the particle will move perpendicular to its initial direction is **20 seconds**.

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