Premed · Premed · Physics 1
Lecture 3: Kinematics in Two Dimensions and Projectile Motion
Physics I — Mechanics & Thermodynamics
Learning Objectives
By the end of this lecture, students will be able to:
- Extend kinematic equations to two-dimensional motion using vector components
- Analyze projectile motion by decomposing it into independent horizontal and vertical components
- Derive and apply equations for range, maximum height, and time of flight
- Solve problems involving projectiles launched at arbitrary angles from arbitrary heights
- Describe relative velocity in two dimensions
Lecture Content
I. Position, Velocity, and Acceleration in 2D
In two dimensions, position is described by a vector r = x i-hat + y j-hat, and displacement becomes Delta r = Delta x i-hat + Delta y j-hat. The velocity vector is v = dx/dt i-hat + dy/dt j-hat = v_x i-hat + v_y j-hat, with speed given by |v| = sqrt(v_x^2 + v_y^2) and direction by theta = arctan(v_y / v_x). Similarly, the acceleration vector is a = dv_x/dt i-hat + dv_y/dt j-hat = a_x i-hat + a_y j-hat.
The key insight for two-dimensional motion is that motion in perpendicular directions is independent. The x-component of motion is governed by a_x alone, and the y-component by a_y alone. Each component obeys its own set of one-dimensional kinematic equations. This independence is what makes two-dimensional problems tractable: they reduce to a pair of one-dimensional problems linked by the shared variable time.
II. Projectile Motion — Setup and Assumptions
A projectile is any object launched into the air that moves under gravity alone after launch. The standard assumptions are that air resistance is negligible, acceleration is constant with a_x = 0 and a_y = -g, Earth's curvature is negligible for short-range projectiles, and g = 9.80 m/s^2 near Earth's surface. Under these assumptions, the trajectory of a projectile is a parabola. For a launch at angle theta_0 with initial speed v_0, the initial velocity components are v_{0x} = v_0 cos(theta_0) and v_{0y} = v_0 sin(theta_0).
III. Equations of Projectile Motion
In the horizontal direction, there is no acceleration, so the motion proceeds at constant velocity: x = x_0 + v_{0x} t = x_0 + (v_0 cos theta_0) t, and v_x = v_{0x} = v_0 cos theta_0 throughout the flight. In the vertical direction, the motion is free fall: y = y_0 + v_{0y} t - (1/2)g t^2 = y_0 + (v_0 sin theta_0) t - (1/2)g t^2, v_y = v_{0y} - g t = v_0 sin theta_0 - g t, and v_y^2 = v_{0y}^2 - 2g(y - y_0). The two sets of equations share the variable t, which serves as the link between horizontal and vertical motion.
<image>A diagram of a projectile launched at angle theta_0 from the origin. The parabolic trajectory is shown with velocity vectors drawn at launch, at the peak, and just before landing. Each velocity vector is decomposed into horizontal (v_x, constant length) and vertical (v_y, changing length and direction) components. The peak is labeled "v_y = 0, v_x = v_{0x}." Axes are labeled x (horizontal) and y (vertical, upward positive). The range R and maximum height H are annotated.</image>
IV. Key Quantities for Symmetric Projectile Motion
For a projectile launched from and landing at the same height (y_0 = 0), several quantities have clean closed-form expressions. The time of flight is T = 2 v_0 sin(theta_0) / g. The maximum height is H = v_0^2 sin^2(theta_0) / (2g), reached at time t = T/2. The range (horizontal distance) is R = v_0^2 sin(2 theta_0) / g. Maximum range occurs at theta_0 = 45 degrees, since sin(90) = 1, and complementary launch angles such as 30 and 60 degrees yield the same range. These formulas are only valid when launch and landing occur at the same elevation. For asymmetric problems, such as launches from a cliff, the full component equations must be used.
V. Trajectory Equation
By eliminating t from the x and y equations, one obtains the trajectory equation that describes the path shape directly: y = x tan(theta_0) - [g / (2 v_0^2 cos^2(theta_0))] x^2. This is the equation of a parabola opening downward. It is particularly useful for finding the height y at a given horizontal position x, or for determining whether a projectile clears an obstacle of known height at a known distance.
VI. Problem-Solving Strategy for Projectiles
A systematic approach to projectile problems begins with choosing a coordinate system, typically with x horizontal and y vertical upward. Next, identify the known quantities (v_0, theta_0, x_0, y_0) and what is being asked. Resolve v_0 into its horizontal and vertical components, then write out the x and y kinematic equations. Use the shared variable t to link the two directions, solve algebraically before substituting numbers, and finally check units, signs, and the reasonableness of the answer.
Common variations include horizontal launch (theta_0 = 0, so v_{0y} = 0 and v_{0x} = v_0), launch from a height where y_0 is not zero, and problems requiring the angle to hit a specific target, which involve trigonometric identities.
<image>Panel A: A horizontal launch from a cliff of height h. The initial velocity is purely horizontal. The trajectory curves downward. Equations for time to hit the ground (t = sqrt(2h/g)) and horizontal distance (x = v_0 t) are annotated. Panel B: Two trajectories from the same launch point with the same speed but complementary angles (30 degrees and 60 degrees). Both land at the same horizontal distance R but the 60-degree trajectory reaches a greater height. The 45-degree maximum-range trajectory is shown as a dashed curve between them.</image>
VII. Relative Velocity in Two Dimensions
The velocity of object A relative to object B is given by v_{A/B} = v_{A} - v_{B}, or equivalently, v_{A} = v_{A/B} + v_{B}. This is a vector addition problem and can be solved using components or graphical methods. Classic examples include a boat crossing a river with a current, an airplane flying in wind, and rain appearing to fall at an angle when viewed from a moving car.
The river-crossing problem illustrates two important optimization strategies. To cross in the shortest time, the boat should aim straight across the river, accepting that the current will carry it downstream. To cross along the shortest path (straight across), the boat should aim upstream at an angle chosen to exactly cancel the current's effect.
<image>A river-crossing diagram showing a boat aiming at angle theta upstream. The river current flows to the right with velocity v_river. The boat's velocity relative to the water points upstream at angle theta. The resultant velocity (boat relative to ground) points straight across the river. Vector triangle showing v_boat/ground = v_boat/water + v_water/ground is drawn beside the river diagram.</image>
