A small particle of mass m is projected at angle Q with the axsis with an intial velocity Vo in X-Y plane at time t

When a small particle of mass \( m \) is projected at an angle \( \theta \) with respect to the horizontal axis in the X-Y plane with an initial velocity \( V_0 \), it follows a specific trajectory influenced by the forces acting on it, primarily gravity. Let’s break down the motion into its components and analyze the situation step by step.

Breaking Down the Motion

The initial velocity \( V_0 \) can be resolved into two components: the horizontal component \( V_{0x} \) and the vertical component \( V_{0y} \). These components can be calculated using trigonometric functions:

• Horizontal Component: \( V_{0x} = V_0 \cdot \cos(\theta) \)
• Vertical Component: \( V_{0y} = V_0 \cdot \sin(\theta) \)

Equations of Motion

Once we have the components of the initial velocity, we can apply the equations of motion to determine the position of the particle at any time \( t \). The equations for the horizontal and vertical positions \( x(t) \) and \( y(t) \) can be expressed as follows:

• Horizontal Position:

  Since there is no acceleration in the horizontal direction (assuming no air resistance), the horizontal position is given by:

  \( x(t) = V_{0x} \cdot t = (V_0 \cdot \cos(\theta)) \cdot t \)

• Vertical Position:

  The vertical motion is influenced by gravity, which acts downwards. The equation for vertical position becomes:

  \( y(t) = V_{0y} \cdot t - \frac{1}{2} g t^2 = (V_0 \cdot \sin(\theta)) \cdot t - \frac{1}{2} g t^2 \)

Understanding the Trajectory

The path traced by the particle is a parabola, characteristic of projectile motion. The maximum height reached by the particle can be found by determining when the vertical velocity becomes zero. This occurs at:

\( t_{max} = \frac{V_{0y}}{g} = \frac{V_0 \cdot \sin(\theta)}{g} \)

Substituting this back into the equation for \( y(t) \) gives the maximum height:

\( H = V_{0y} \cdot t_{max} - \frac{1}{2} g t_{max}^2 \)

Range of the Projectile

The total time of flight until the particle returns to the same vertical level can be calculated as:

\( T = \frac{2 V_{0y}}{g} = \frac{2 V_0 \cdot \sin(\theta)}{g} \)

The horizontal range \( R \) of the projectile can then be found by substituting \( T \) into the horizontal position equation:

\( R = V_{0x} \cdot T = (V_0 \cdot \cos(\theta)) \cdot \left(\frac{2 V_0 \cdot \sin(\theta)}{g}\right) \)

Thus, the range becomes:

\( R = \frac{V_0^2 \cdot \sin(2\theta)}{g} \)

Real-World Applications

This understanding of projectile motion has numerous applications, from sports (like basketball or soccer) to engineering (such as calculating the trajectory of missiles or fireworks). By manipulating the angle and initial velocity, one can optimize the distance or height achieved by the projectile.

In summary, analyzing the motion of a particle projected at an angle involves breaking down its velocity into components, applying the equations of motion, and understanding the resulting trajectory. This foundational knowledge is crucial in both theoretical physics and practical applications. If you have any further questions or need clarification on specific aspects, feel free to ask!