A particle of mass 2kg moves with an initial velocity of `vec(v)=4hat(i) +4hat(j) ms^(-1)`. A constant force of `vec(F) =20hat(j)N` is applied on the particle. Initially, the particle was at (0,0). The x-coordinates of the particle when its y-coordinates again becomes zero is given by

Options

A. 3.2m
B. 6m
C. 4.8m
D. 1.2m

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the influence of the applied force and find the x-coordinate when the y-coordinate returns to zero.

### Step 1: Identify the initial conditions
The particle has:
- Mass \( m = 2 \, \text{kg} \)
- Initial velocity \( \vec{v} = 4 \hat{i} + 4 \hat{j} \, \text{m/s} \)
- \( v_x = 4 \, \text{m/s} \)
- \( v_y = 4 \, \text{m/s} \)
- Initial position \( (0, 0) \)

### Step 2: Determine the effect of the applied force
A constant force \( \vec{F} = 20 \hat{j} \, \text{N} \) is applied. This force will only affect the y-component of the motion since it acts in the vertical direction.

### Step 3: Calculate the acceleration due to the force
Using Newton's second law, \( \vec{F} = m \vec{a} \):
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{20 \hat{j}}{2} = 10 \hat{j} \, \text{m/s}^2 \]
Thus, the acceleration in the y-direction is \( a_y = 10 \, \text{m/s}^2 \) and there is no acceleration in the x-direction (\( a_x = 0 \)).

### Step 4: Determine the time taken to return to y = 0
The y-coordinate of the particle can be described by the equation of motion:
\[ y(t) = y_0 + v_{y0} t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ y(t) = 0 + 4t + \frac{1}{2} (10) t^2 \]
\[ y(t) = 4t + 5t^2 \]
To find when the particle returns to \( y = 0 \):
\[ 4t + 5t^2 = 0 \]
Factoring out \( t \):
\[ t(5t + 4) = 0 \]
This gives us two solutions:
1. \( t = 0 \) (initial position)
2. \( 5t + 4 = 0 \) leading to \( t = -\frac{4}{5} \) (not valid)
3. The other solution is \( t = 0 \) and we need to find the time when it returns to zero after some time.

### Step 5: Find the time when y-coordinate is zero again
The correct approach is to consider the motion until it reaches the maximum height and then returns to y = 0. To find the time to reach the maximum height, we set the vertical velocity to zero:
\[ v_y = v_{y0} + a_y t \]
Setting \( v_y = 0 \):
\[ 0 = 4 + 10t \implies t = -\frac{4}{10} = -0.4 \, \text{s} \text{ (not valid)} \]
Instead, we can use the quadratic formula to find the time when the particle returns to y = 0.

### Step 6: Calculate the range using the formula
The range \( R \) of a projectile is given by:
\[ R = \frac{2 v_x v_y}{g} \]
Where \( g = 10 \, \text{m/s}^2 \):
Substituting the values:
\[ R = \frac{2 \cdot 4 \cdot 4}{10} = \frac{32}{10} = 3.2 \, \text{m} \]

### Conclusion
The x-coordinate of the particle when its y-coordinate becomes zero again is \( 3.2 \, \text{m} \).