What is nature of trajectory of a particle having a uniformly accelerated motion in a plane?

Answer

To determine the nature of the trajectory of a particle undergoing uniformly accelerated motion in a plane, we can analyze the motion by breaking it down into its components along the x and y axes. Here’s a step-by-step solution:

### Step 1: Understand the Motion
The particle is moving in a two-dimensional plane under the influence of gravity. It has an initial velocity \( u \) at an angle \( \theta \) with respect to the horizontal axis.

**Hint:** Identify the axes and the initial conditions of the motion.

### Step 2: Break Down the Initial Velocity
The initial velocity can be resolved into two components:
- Horizontal component: \( u_x = u \cos(\theta) \)
- Vertical component: \( u_y = u \sin(\theta) \)

**Hint:** Use trigonometric functions to resolve the velocity into components.

### Step 3: Analyze the Accelerations
In the horizontal direction (x-axis), there is no acceleration (i.e., \( a_x = 0 \)). In the vertical direction (y-axis), the only acceleration acting on the particle is due to gravity, which we denote as \( a_y = -g \) (negative because it acts downward).

**Hint:** Recognize that the horizontal motion is uniform while the vertical motion is uniformly accelerated.

### Step 4: Write the Equations of Motion
For the horizontal motion, the distance \( x \) covered can be expressed as:
\[ x = u_x \cdot t = u \cos(\theta) \cdot t \]
From this, we can express time \( t \) as:
\[ t = \frac{x}{u \cos(\theta)} \]

For the vertical motion, we can use the second equation of motion:
\[ y = u_y \cdot t + \frac{1}{2} a_y \cdot t^2 \]
Substituting for \( u_y \) and \( a_y \):
\[ y = u \sin(\theta) \cdot t - \frac{1}{2} g t^2 \]

**Hint:** Use the equations of motion to relate distance, time, and acceleration.

### Step 5: Substitute Time into the Vertical Motion Equation
Now substitute the expression for \( t \) from the horizontal motion into the vertical motion equation:
\[ y = u \sin(\theta) \cdot \left(\frac{x}{u \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{x}{u \cos(\theta)}\right)^2 \]

### Step 6: Simplify the Equation
This simplifies to:
\[ y = x \tan(\theta) - \frac{g}{2u^2 \cos^2(\theta)} x^2 \]

### Step 7: Identify the Nature of the Trajectory
The equation \( y = mx - kx^2 \) (where \( m = \tan(\theta) \) and \( k = \frac{g}{2u^2 \cos^2(\theta)} \)) represents a quadratic equation in \( x \). This indicates that the trajectory of the particle is a parabola.

**Hint:** Recognize the form of the equation to determine the nature of the trajectory.

### Conclusion
The nature of the trajectory of a particle having uniformly accelerated motion in a plane is a **parabola**.