Kinematics in Two Dimensions: Projectile Motion Study Guide

Motion study has progressed from simple straight-line motion (linear motion) to nonlinear motion, which involves movement along a curved path.

Projectile motion involves an object moving in two dimensions under the exclusive influence of Earth's gravity.
A projectile is the specific object that is thrown or launched into the air.
A trajectory is the term for the path followed by the projectile during its flight.

Horizontal Projectile Motion: This occurs when an object is launched horizontally from a specific height.
Vertical Projectile Motion: This occurs when an object is launched at an angle relative to the horizontal.

Because a projectile moves in two dimensions, it functions like a resultant vector and possesses two distinct components: horizontal and vertical.
For calculation purposes, the horizontal and vertical parts of the motion are considered separately.
A fundamental assumption in these models is that air resistance is negligible.

Comparative Motion Observations:
- A multiple-exposure photograph of two balls—one dropped from rest and the other projected horizontally outward at the same time—reveals that their vertical positions remain identical at every instant.
Vertical Velocity Component ( $y$ -component):
- The vertical velocity changes due to the influence of gravity.
- The object does not cover equal vertical displacements in equal time periods; the vertical distance covered increases with every subsequent second.
- The vertical acceleration is constant and defined as $a_y = -g$ .
- For a horizontally launched projectile, the initial vertical velocity is zero ( $v_{0y} = 0$ ), but this velocity increases continually in the downward direction until the object reaches the ground.
Horizontal Velocity Component ( $x$ -component):
- The horizontal velocity never changes and covers equal displacements in equal time periods.
- There is no influence of gravity or any other acceleration in the horizontal direction ( $a_x = 0$ ).
- The horizontal component of velocity ( $v_x$ ) remains constant throughout the flight and is equal to its initial value ( $v_{0x}$ ).

Horizontal ( $x$ ) Component:
- Magnitude: Constant.
- Direction: Constant.
Vertical ( $y$ ) Component:
- Magnitude: Changes over time.
- Direction: Changes over time.

Horizontal ( $x$ ) Component Equations:
- $v_x = v_{0x} + a_xt$
- $x = \frac{1}{2}(v_{0x} + v_x)t$
- $x = v_{0x}t + \frac{1}{2}a_xt^2$
- $v_x^2 = v_{0x}^2 + 2a_xx$
Vertical ( $y$ ) Component Equations:
- $v_y = v_{0y} + a_yt$
- $y = \frac{1}{2}(v_{0y} + v_y)t$
- $y = v_{0y}t + \frac{1}{2}a_yt^2$
- $v_y^2 = v_{0y}^2 + 2a_yy$

When simplified for projectile motion where $a_x = 0$ and $a_y = -g$ :
- $v_x = v_{0x}$
- $x = x_0 + v_{0x}t$
- $v_y = v_{0y} - gt$
- $y = y_0 + v_{0y}t - \frac{1}{2}gt^2$

Scenario: A plane traveling with a horizontal velocity of $100. m/s$ is at an altitude of $500. m$ above the ground. The pilot drops a bomb on a target.
Given Data:
- Horizontal velocity ( $v_{0x}$ ): $100. m/s$
- Initial vertical position ( $y_0$ ): $0.00 m$
- Vertical displacement ( $y$ ): $500. m$
- Initial vertical velocity ( $v_{0y}$ ): $0.00 m/s$
- Vertical acceleration ( $a_y$ ): $-9.80 m/s^2$
Problem (a): Calculation of time ( $t$ ) the bomb is in the air:
- Equation: $y = v_{0y}t - \frac{1}{2}gt^{2}$
- Rearranged for time: $t = \sqrt{\frac{2y}{g}}$
- Solution: $t = \sqrt{\frac{2(500. m)}{9.80 m/s^2}} = 10.1 s$
Problem (b): Calculation of horizontal distance ( $x$ ) from the release point to the impact point:
- Equation: $x = v_{0x}t$
- Solution: $x = (100. m/s)(10.10 s) = 1010 m$

Velocity Behavior:
- Horizontal Velocity: Remains constant throughout the trajectory.
- Vertical Velocity: Decreases as the object moves upward, reaches exactly zero at the top of the trajectory, and then increases as the object moves downward.
Summary of Properties:
- Horizontal ( $x$ ) Component: Magnitude is constant; direction is constant.
- Vertical ( $y$ ) Component: Magnitude decreases on the way up, is $0$ at the top, and increases on the way down; direction changes.

Projectiles launched at an angle ( $\theta$ ) must have their initial velocity ( $v_0$ ) broken into components using trigonometry:
- Horizontal Component: $v_{0x} = v_0 \times \text{cos}(\theta)$
- Vertical Component: $v_{0y} = v_0 \times \text{sin}(\theta)$
Ground-to-Ground Logic: If a projectile begins and ends its flight at ground level, the total vertical displacement ( $y$ ) is zero ( $y = 0$ ).

Scenario: A place kicker kicks a football with an initial velocity of $20.0 m/s$ at an angle of $53.0^{\circ}$ .
Given Data:
- Initial Velocity ( $v_0$ ): $20.0 m/s$
- Angle ( $\theta$ ): $53.0^{\circ}$
- Initial Vertical Position ( $y_0$ ): $0.00 m$
Step 1: Resolve Velocity Components:
- $v_{0x} = (20.0 m/s) \times \text{cos}(53.0^{\circ}) = 12.0 m/s$
- $v_{0y} = (20.0 m/s) \times \text{sin}(53.0^{\circ}) = 16.0 m/s$
Problem (a): Calculation of total time in the air ( $t$ ):
- Equation: $y = v_{0y}t - \frac{1}{2}gt^{2}$
- Setting $y = 0$ for ground-to-ground: $0 = (16.0 m/s)t - (4.90 m/s^2)t^2$
- Dividing by $t$ : $16.0 m/s = 4.90 m/s^2 \times t$
- Solution: $t = 3.27 s$
Problem (b): Calculation of horizontal range ( $x$ ):
- Equation: $x = x_0 + v_{0x}t$
- Solution: $x = 0 m + (12.0 m/s)(3.27 s) = 39.2 m$
Problem (c): Calculation of maximum height reached ( $y_{\text{max}}$ ):
- Principle: The time to reach maximum height is exactly half of the total flight time for a symmetric trajectory ( $t_{\text{to reach } y_{\text{max}}} = \frac{3.27 s}{2} = 1.64 s$ ).
- Equation: $y_{\text{max}} = v_{0y}t - \frac{1}{2}gt^2$
- Solution: $y_{\text{max}} = (16.0 m/s)(1.64 s) - \frac{1}{2}(9.80 m/s^2)(1.64 s)^{2}$
- Final Answer: $y_{\text{max}} = 13.1 m$