Motion in Two Dimensions: Kinematics and Projectile Motion

Motion in Two Dimensions

Equations for 2-D Kinematics at Constant Acceleration

Two-dimensional kinematics describes the motion of objects in a plane, using vector components for position, velocity, and acceleration. When acceleration is constant, the following equations apply to each direction independently:

• Position as a function of time:

  • $x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2$

  • $y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2$

• Velocity as a function of time:

  • $v_x = v_{0x} + a_x t$

  • $v_y = v_{0y} + a_y t$

• Velocity as a function of displacement:

  • $v_x^2 = v_{0x}^2 + 2 a_x (x - x_0)$

  • $v_y^2 = v_{0y}^2 + 2 a_y (y - y_0)$

Key Terms: Displacement, velocity, acceleration, vector components

Example: Calculating the position and velocity of a car moving in a straight line with constant acceleration.

Projectile Motion

Projectile motion occurs when an object is launched into the air and moves under the influence of gravity alone (free fall). The acceleration due to gravity is a constant vector, $g$, directed downward.

• Horizontal acceleration: $a_x = 0$

• Vertical acceleration: $a_y = -g$

• Horizontal velocity: $v_x = v_{0x}$ (constant)

• Vertical velocity: $v_y = v_{0y} - g t$

Trajectory equations:

• $x = x_0 + v_{0x} t$

• $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$

Example: A ball thrown horizontally from a cliff follows a parabolic path due to gravity.

Free-Fall Trajectory

When an object is projected from the origin $(x_0 = 0, y_0 = 0)$ with initial velocity $v_0$ at an angle $\theta$, its trajectory is described by:

• $x = v_0 \cos \theta \; t$

• $y = v_0 \sin \theta \; t - \frac{1}{2} g t^2$

Key Point: The path is a parabola, and the horizontal and vertical motions are independent except for sharing the same time variable.

Problem Solving: Example of Soccer Ball Projectile

Consider a child kicking a soccer ball from ground level with initial speed $V_0$ at angle $\theta$ to the horizontal. The ball hits a wall a distance $L$ away.

• a) Diagram: The initial velocity vector $V_0$ is resolved into horizontal ($V_0 \cos \theta$) and vertical ($V_0 \sin \theta$) components. The wall is at $x = L$.

• b) Time to reach the wall: Set $x = L$ and solve for $t$:

  • $L = V_0 \cos \theta \; t$

  • $\Rightarrow t = \frac{L}{V_0 \cos \theta}$

• c) Height at which the ball hits the wall: Substitute $t$ into the $y$ equation:

  • $H = V_0 \sin \theta \; t - \frac{1}{2} g t^2$

  • Using $t$ from above:

  • $H = V_0 \sin \theta \left( \frac{L}{V_0 \cos \theta} \right) - \frac{1}{2} g \left( \frac{L}{V_0 \cos \theta} \right)^2$

  • $H = L \tan \theta - \frac{g L^2}{2 V_0^2 \cos^2 \theta}$

• d) Maximum height above the ground: The maximum height occurs when $v_y = 0$:

  • $v_y = V_0 \sin \theta - g t_{max} = 0$

  • $t_{max} = \frac{V_0 \sin \theta}{g}$

  • Substitute $t_{max}$ into $y$:

  • $H_{max} = V_0 \sin \theta \left( \frac{V_0 \sin \theta}{g} \right) - \frac{1}{2} g \left( \frac{V_0 \sin \theta}{g} \right)^2$

  • $H_{max} = \frac{V_0^2 \sin^2 \theta}{2g}$

Additional info: These symbolic expressions are fundamental for analyzing projectile motion in physics, and can be adapted for different initial conditions or target locations.

Example: Range of a Projectile

The range $R$ of a projectile launched from ground level is the horizontal distance it travels before returning to the same vertical position ($y = 0$):

• $R = \frac{V_0^2 \sin 2\theta}{g}$

Key Point: The range is maximized when $\theta = 45^\circ$ for a given initial speed.

Demo: The Hunter and the Monkey

This classic physics demonstration illustrates that a projectile aimed directly at a falling target will hit the target, provided both experience the same gravitational acceleration. The horizontal and vertical motions are independent, but gravity affects both equally.

• Application: Predicting the intersection point of two objects in free fall.