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, we need to determine the time at which the velocity of the particle makes a 45-degree angle with its initial direction of projection. Here's the step-by-step solution:

### Step 1: Understand the motion of the particle
The particle is projected horizontally with an initial speed of \( u = 20 \, \text{m/s} \). Since it is projected horizontally, its initial vertical velocity \( v_{y0} = 0 \).

### Step 2: Analyze horizontal and vertical motions
- The horizontal velocity \( v_x \) remains constant because there is no horizontal acceleration:
\[ v_x = 20 \, \text{m/s} \]

- The vertical velocity \( v_y \) increases due to gravitational acceleration \( g \):
\[ v_y = g \cdot t \]
where \( g \) is approximately \( 10 \, \text{m/s}^2 \) (taking the value of gravitational acceleration).

### Step 3: Set up the condition for a 45-degree angle
For the resultant velocity to make a 45-degree angle with the horizontal, the magnitudes of the horizontal and vertical velocities must be equal:
\[ v_y = v_x \]

Substituting the expressions we have:
\[ g \cdot t = 20 \, \text{m/s} \]

### Step 4: Solve for time \( t \)
Using \( g = 10 \, \text{m/s}^2 \):
\[ 10 \cdot t = 20 \]
\[ t = \frac{20}{10} = 2 \, \text{s} \]

### Final Answer
The time after which the velocity of the particle will be at a 45-degree angle from the initial direction of projection is \( t = 2 \, \text{s} \).

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