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. `(ba)/(c )`
B. `(2ba)/(c )`
C. `(3ba)/(c )`
D. None

Explanation

To find the range of the particle projected from the horizontal plane (the x-z plane) with the given velocity vector \(\vec{V} = a \hat{i} + (b - ct) \hat{j}\), we can follow these steps:

### Step 1: Identify the components of the velocity vector
The velocity vector is given as:
\[ \vec{V} = a \hat{i} + (b - ct) \hat{j} \]
From this, we can identify the horizontal and vertical components of the velocity:
- Horizontal component (\(V_x\)): \(a\)
- Vertical component (\(V_y\)): \(b - ct\)

### Step 2: Determine the time of flight
The vertical motion is affected by the downward acceleration due to gravity. In this case, the downward acceleration is represented by \(-c\). The vertical velocity can be expressed as:
\[ V_y = b - ct \]
To find the time of flight, we need to determine when the particle returns to the horizontal plane (i.e., when \(y = 0\)). The vertical position \(y\) can be found by integrating the vertical velocity:
\[ y(t) = \int (b - ct) dt = bt - \frac{1}{2}ct^2 + y_0 \]
Assuming the particle starts from the horizontal plane (\(y_0 = 0\)), we set \(y(t) = 0\) to find the time of flight:
\[ 0 = bt - \frac{1}{2}ct^2 \]
Factoring out \(t\):
\[ t(bt - \frac{1}{2}ct) = 0 \]
This gives us two solutions:
1. \(t = 0\) (initial time)
2. \(bt - \frac{1}{2}ct = 0 \Rightarrow t = \frac{2b}{c}\)

### Step 3: Calculate the horizontal range
The horizontal range \(R\) can be calculated using the horizontal component of the velocity and the total time of flight:
\[ R = V_x \cdot t_{total} \]
Substituting the values:
\[ R = a \cdot \frac{2b}{c} \]
Thus, the range \(R\) is given by:
\[ R = \frac{2ab}{c} \]

### Final Answer
The range on the horizontal plane is:
\[ R = \frac{2ab}{c} \]

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