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. 1.2m
B. 4.8m
C. 6.0m
D. 3.2m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given quantities
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Initial velocity, \( \vec{v} = 4\hat{i} + 4\hat{j} \, \text{m/s} \)
- Force applied, \( \vec{F} = 20\hat{j} \, \text{N} \)

### Step 2: Calculate the acceleration
Using Newton's second law, \( \vec{F} = m\vec{a} \), we can find the acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{20\hat{j}}{2} = 10\hat{j} \, \text{m/s}^2 \]
This means the acceleration in the y-direction is \( 10 \, \text{m/s}^2 \) upward.

### Step 3: Write the equations of motion for the y-coordinate
The equation for the y-coordinate is given by:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Where:
- \( u_y = 4 \, \text{m/s} \) (initial velocity in the y-direction)
- \( a_y = 10 \, \text{m/s}^2 \) (acceleration in the y-direction)

Substituting the values:
\[ y = 4t + \frac{1}{2} \cdot 10 \cdot t^2 = 4t + 5t^2 \]

### Step 4: Set the y-coordinate to zero and solve for time
We want to find when the y-coordinate becomes zero:
\[ 0 = 4t + 5t^2 \]
Factoring out \( t \):
\[ t(5t + 4) = 0 \]
This gives us two solutions:
1. \( t = 0 \) (initial position)
2. \( 5t + 4 = 0 \Rightarrow t = -\frac{4}{5} \) (not physically meaningful)
3. The other solution is \( t = 0 \) and we need to find the time when it returns to \( y = 0 \).

### Step 5: Find the time when the particle returns to y=0
Rearranging the equation:
\[ 5t^2 + 4t = 0 \]
The valid solution is:
\[ t = \frac{-4}{5} \text{ (not valid)}, \text{ we need to find the time when it reaches back to } y = 0. \]

### Step 6: Use the quadratic formula
The equation \( 5t^2 + 4t = 0 \) can be solved using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5, b = 4, c = 0 \):
\[ t = \frac{-4 \pm \sqrt{(4)^2 - 4 \cdot 5 \cdot 0}}{2 \cdot 5} = \frac{-4 \pm 4}{10} \]
This gives us:
1. \( t = 0 \)
2. \( t = -\frac{8}{10} = -0.8 \text{ (not valid)} \)

### Step 7: Find the x-coordinate when y=0
Since there is no acceleration in the x-direction, the x-coordinate can be calculated as:
\[ x = u_x t \]
Where \( u_x = 4 \, \text{m/s} \) and \( t = 0.8 \, \text{s} \):
\[ x = 4 \cdot \frac{4}{5} = \frac{16}{5} = 3.2 \, \text{m} \]

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

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