Particle is dropped form the height of `20m` from horizontal ground. There is wind blowing due to which horizontal acceleration of the particles becomes `6 ms^(-2)`. Find the horizontal displacement of the particle till it reaches ground.

Options

A. 6m
B. 10m
C. 12m
D. 24m

Explanation

To solve the problem of finding the horizontal displacement of a particle dropped from a height of 20 meters with a horizontal acceleration of 6 m/s², we can follow these steps:

### Step-by-Step Solution:

1. **Determine the time taken to fall**:
The time \( t \) taken for the particle to fall from a height \( h \) can be calculated using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Given:
- Height \( h = 20 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting the values:
\[ t = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2 \, \text{s} \]

2. **Calculate the horizontal displacement**:
The horizontal displacement \( R \) can be calculated using the formula:
\[ R = v_x \cdot t + \frac{1}{2} a_x t^2 \]
where:
- \( v_x \) is the initial horizontal velocity (which is \( 0 \) since the particle is dropped),
- \( a_x \) is the horizontal acceleration (given as \( 6 \, \text{m/s}^2 \)),
- \( t \) is the time calculated in the previous step.

Since \( v_x = 0 \):
\[ R = 0 \cdot t + \frac{1}{2} \cdot 6 \cdot (2)^2 \]
Simplifying this:
\[ R = 0 + \frac{1}{2} \cdot 6 \cdot 4 = 12 \, \text{m} \]

3. **Final Result**:
The horizontal displacement of the particle until it reaches the ground is \( 12 \, \text{m} \).

### Summary:
The horizontal displacement of the particle is \( 12 \, \text{m} \).