A particle is projected form a horizontal plane `(x-z` plane) such that its velocity vector at time `t` is gives by `vecV = ahati +(b - ct)hatj`. Its range on the horizontal plane is given by

Options

A. `2ab//c`
B. `ab//c`
C. `ac//b`
D. `a//2bc`

Explanation

To solve the problem of finding the range of a particle projected from a horizontal plane with a given velocity vector, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Velocity Components**:
The velocity vector is given as \(\vec{V} = a \hat{i} + (b - ct) \hat{j}\). Here, we can identify the horizontal and vertical components of the velocity:
- Horizontal component (\(u_x\)): \(a\)
- Vertical component (\(u_y\)): \(b - ct\)

2. **Understand the Motion**:
The particle is projected in a projectile motion. The horizontal motion is uniform (constant velocity), while the vertical motion is affected by gravity.

3. **Determine the Initial Velocities**:
At time \(t = 0\):
- \(u_x = a\)
- \(u_y = b\)

4. **Calculate the Range Formula**:
The range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

5. **Express \(\sin(2\theta)\)**:
We can express \(\sin(2\theta)\) as:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
In terms of the components:
\[ \sin(\theta) = \frac{u_y}{u} \quad \text{and} \quad \cos(\theta) = \frac{u_x}{u} \]

6. **Calculate the Magnitude of Initial Velocity**:
The magnitude of the initial velocity \(u\) can be calculated as:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{a^2 + b^2} \]

7. **Substitute into the Range Formula**:
Now, substituting \(u_x\) and \(u_y\) into the range formula:
\[ R = \frac{(u_x^2 + u_y^2) \cdot 2 \cdot u_x \cdot u_y}{g} \]
Here, \(g\) is given as \(c\) (the negative acceleration due to gravity).

8. **Final Expression for Range**:
Substituting the values:
\[ R = \frac{2 \cdot a \cdot b}{c} \]

### Final Answer:
Thus, the range of the particle on the horizontal plane is given by:
\[ R = \frac{2ab}{c} \]