A force `barF=barVxxbarA` is exerted on a particle in addition to the force of gravity, where `barV` is the velocity of the particle and `barA` is a constant vector in the horizontal direction. The minimum speed of a particle of mass 3kg be projected, so that it continues to move undeflected with a constant velocity is `(xg)/A.` Then 'x' is (consider the plane of the ground as xy plane)

Answer

To solve the problem, we need to analyze the forces acting on the particle and derive the minimum speed required for it to move undeflected with constant velocity.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
- The gravitational force acting downward on the particle is given by \( \bar{F}_g = m\bar{g} \), where \( m = 3 \, \text{kg} \) is the mass of the particle and \( \bar{g} \) is the acceleration due to gravity.
- An additional force \( \bar{F} = \bar{V} \times \bar{A} \) is acting on the particle, where \( \bar{V} \) is the velocity of the particle and \( \bar{A} \) is a constant vector in the horizontal direction.

2. **Set Up the Equation for Constant Velocity:**
- For the particle to move with constant velocity, the net force acting on it must be zero. This means that the upward component of the force \( \bar{F} \) must balance the downward gravitational force \( \bar{F}_g \).
- The upward component of the force can be expressed as \( F_{\text{up}} = |\bar{V}| |\bar{A}| \sin \theta \), where \( \theta \) is the angle between the velocity vector \( \bar{V} \) and the vector \( \bar{A} \).

3. **Equate the Forces:**
- Since the net force is zero, we can write:
\[ |\bar{V}| |\bar{A}| \sin \theta = mg \]
- Rearranging gives us:
\[ |\bar{V}| = \frac{mg}{|\bar{A}| \sin \theta} \]

4. **Determine Minimum Speed:**
- To find the minimum speed \( |\bar{V}| \), we need to maximize \( \sin \theta \). The maximum value of \( \sin \theta \) is 1 (when \( \theta = 90^\circ \)).
- Thus, the minimum speed becomes:
\[ |\bar{V}|_{\text{min}} = \frac{mg}{|\bar{A}|} \]

5. **Substitute Known Values:**
- Given \( m = 3 \, \text{kg} \) and substituting \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)):
\[ |\bar{V}|_{\text{min}} = \frac{3 \times 9.81}{|\bar{A}|} = \frac{29.43}{|\bar{A}|} \]

6. **Compare with Given Expression:**
- The problem states that the minimum speed is given by \( \frac{xg}{A} \).
- We can equate this with our derived expression:
\[ \frac{29.43}{|\bar{A}|} = \frac{xg}{A} \]
- Since \( g \) is a constant, we can compare coefficients:
\[ x = 3 \]

### Final Answer:
Thus, the value of \( x \) is \( 3 \).

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