A particle is projected horizontally will speed `20 ms^(-1)` from the top of a tower. After what time velocity of particle will be at `45^(@)` angle from initial direction of projection.

Options

A. 1s
B. 2s
C. 3s
D. 4s

Explanation

To solve the problem of a particle projected horizontally from the top of a tower with a speed of \(20 \, \text{m/s}\), we need to determine the time at which the velocity of the particle makes a \(45^\circ\) angle with the horizontal direction of projection.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The particle is projected horizontally with an initial horizontal velocity \(v_x = 20 \, \text{m/s}\).
- The initial vertical velocity \(v_y\) is \(0 \, \text{m/s}\) since it is projected horizontally.

2. **Identify the Components of Velocity**:
- The horizontal component of velocity remains constant:
\[ v_x = 20 \, \text{m/s} \]
- The vertical component of velocity changes due to gravity:
\[ v_y = g \cdot t \]
where \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

3. **Condition for \(45^\circ\) Angle**:
- For the velocity vector to make a \(45^\circ\) angle with the horizontal, the magnitudes of the horizontal and vertical components must be equal:
\[ v_y = v_x \]

4. **Set Up the Equation**:
- Substitute the expressions for \(v_y\) and \(v_x\):
\[ g \cdot t = 20 \]

5. **Solve for Time \(t\)**:
- Rearranging the equation gives:
\[ t = \frac{20}{g} \]
- Substituting \(g = 10 \, \text{m/s}^2\):
\[ t = \frac{20}{10} = 2 \, \text{s} \]

6. **Conclusion**:
- The time after which the velocity of the particle will be at a \(45^\circ\) angle from the initial direction of projection is \(2 \, \text{seconds}\).

### Final Answer:
The time is \(t = 2 \, \text{seconds}\).