Notes on 2D Motion and Projectile Motion

2D Motion and Projectile Motion — Study Notes

• Learning objectives (overall):

  • Extend the definitions of position, velocity, and acceleration to 2D and 3D using vectors.

  • Understand the independence of vertical and horizontal components in projectile motion.

  • Calculate range, time of flight, and maximum height for projectiles.

  • Describe motion using relative velocities in 1D and 2D.

• Quick context from the slides:

  • Topics include 2D motion, projectile motion, vector algebra, average vs instantaneous quantities, and problem-solving strategies.

  • Worked examples cover turtle motion, rover motion on Mars, and 2D projectile problems.

  • Several classic projectile-motion results are highlighted (range, height, trajectory shape).

Vector representation and components

• A vector A has magnitude A and direction; components are the legs of the right triangle:

  • $A<em>x = A \, \cos\theta, \qquad A</em>y = A \ sin\theta,$

  • $A = \sqrt{A<em>x^2 + A</em>y^2}, \qquad \theta = \tan^{-1}\left( rac{A<em>y}{A</em>x}\right).$

• In 2D, vectors are written as

  • $\mathbf{A} = A<em>x \hat{\mathbf{i}} + A</em>y \hat{\mathbf{j}}.$

• Position, displacement, and velocity can be treated as vectors in the plane:

  • Position: $\mathbf{r}(t) = x(t)\hat{\mathbf{i}} + y(t)\hat{\mathbf{j}}.$

  • Displacement between two instants: $\Delta\mathbf{r} = \mathbf{r}(t<em>2) - \mathbf{r}(t</em>1) = \Delta x\hat{\mathbf{i}} + \Delta y\hat{\mathbf{j}}.$

• Average velocity and instantaneous velocity in 2D:

  • $\mathbf{v}_{\text{avg}} = \frac{\Delta \mathbf{r}}{\Delta t}$ in the limit as $\Delta t \to 0$:

  • $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left(\frac{dx}{dt}\right)\hat{\mathbf{i}} + \left(\frac{dy}{dt}\right)\hat{\mathbf{j}}.$

• Acceleration in 2D:

  • $\mathbf{a} = \frac{d\mathbf{v}}{dt} = \left(\frac{d^2x}{dt^2}\right)\hat{\mathbf{i}} + \left(\frac{d^2y}{dt^2}\right)\hat{\mathbf{j}}.$

2D motion: independent components

• In two (or three) dimensions, horizontal and vertical motions are independent:

  • Horizontal motion: $x(t) = x<em>0 + v</em>{0x} t + \frac{1}{2} a_x t^2,$

  • Vertical motion: $y(t) = y<em>0 + v</em>{0y} t + \frac{1}{2} a_y t^2,$

  • with the corresponding velocities $v<em>x(t) = v</em>{0x} + a<em>x t, \quad v</em>y(t) = v<em>{0y} + a</em>y t.$

• For projectile motion near Earth without air resistance: typically

  • $a_x = 0 \, (\text{constant velocity in x}),$

  • $a_y = -g, \quad g \approx 9.8\ \text{m/s}^2.$

• The time $t$ is the same for both x and y directions; projection coordinates are linked by the shared time variable.

Motion in two dimensions: key results

• Trajectory is a parabola: express y as a function of x (trajectory equation):

  • If initial speed is $v0$ and launch angle is $\theta0$:

  • $y(x) = x \tan\theta<em>0 - \frac{g x^2}{2 v</em>0^2 \cos^2\theta_0}.$

• Initial conditions for projectile problems:

  • $x<em>0 = 0,\quad y</em>0 = 0,$

  • $v<em>{0x} = v</em>0 \cos\theta<em>0, \quad v</em>{0y} = v<em>0 \sin\theta</em>0,$

  • Horizontal acceleration $ax = 0$, vertical acceleration $ay = -g$.

• Horizontal and vertical motions are independent; time $t$ is the same for both.

• Key results derived from constant-acceleration in each dimension:

  • Horizontal motion: $x = x<em>0 + v</em>{0x} t,$ (if $a_x=0$)

  • Vertical motion: $y = y<em>0 + v</em>{0y} t - frac{1}{2} g t^2,$

  • Vertical velocity: $v<em>y(t) = v</em>{0y} - g t.$

• Projectile quantities of interest:

  • Time of flight (to return to same height): $T = \frac{2 v_0 \sin\theta}{g}.$

  • Horizontal range (range on level ground): $R = \frac{v_0^2 \sin(2\theta)}{g}.$

  • Maximum height: $h<em>{\max} = \frac{v</em>0^2 \sin^2\theta}{2 g}.$

  • Peak time: $t<em>{\text{peak}} = \frac{v</em>0 \sin\theta}{g}.$

• Trajectory in terms of x: the same-range property for complementary angles: angles $\theta$ and $90^{\circ}-\theta$ give the same range, though heights differ.

• Velocity and speed during flight:

  • Instantaneous velocity components: $v<em>x(t) = v</em>0 \cos\theta,$ (for idealized constant $v{0x}$) and $v</em>y(t) = v_0 \sin\theta - g t.$

  • The speed magnitude: $|\mathbf{v}(t)| = \sqrt{v<em>x(t)^2 + v</em>y(t)^2}.$

  • The direction of motion changes as the vertical component evolves under gravity.

Examples and problem-solving approaches

• Example: Motion of a turtle (2D kinematics with components)

  • Given: initial speed $v_0 = 10\ \text{cm/s}$ at an angle of $25^{\circ}$ to the horizontal.

  • Components:

  • $v<em>{0x} = v</em>0 \cos 25^{\circ} \approx 9.06\ \text{cm/s},$

  • $v<em>{0y} = v</em>0 \sin 25^{\circ} \approx 4.23\ \text{cm/s}.$

  • After time $t$, horizontal displacement: $\Delta x = v_{0x} t.$

  • After time $t$, vertical displacement: $\Delta y = v_{0y} t - \tfrac{1}{2} g t^2$ (with $g \approx 9.8\ \text{m/s}^2$; unit consistency matters).

  • Example numerical illustration (for pedagogy): at $t=10\ \text{s}$, the horizontal displacement would be $\Delta x \approx 0.906\ \text{m}$ (using $v{0x}=0.0906\ \text{m/s}$ if $v0=10\ \text{cm/s}$), while vertical displacement would be large negative due to gravity; the exact numbers depend on unit consistency (cm vs m).

  • Instantaneous velocity components at time $t$: $vx(t)=v{0x}$ (for constant $ax=0$) and $vy(t)=v_{0y}-g t$.

• Example: Rover on Mars (problem setup from slides)

  • The rover’s coordinates vary with time as:

  • $x(t) = 2.0\ \text{m} + (0.25\ \text{m/s}^2)\, t^2,$

  • $y(t) = (1.0\ \text{m/s})\, t + (0.025\ \text{m/s}^3)\, t^3.$

  • (a) Find the rover’s coordinates and distance from the lander at $t=2.0\ \text{s}$.

  • (b) Find the rover’s displacement and average velocity over $t: 0 \to 2.0\ \text{s}.$

  • (c) Find a general expression for the instantaneous velocity vector $\oldsymbol{V}(t)$ and express $t=2.0\ \text{s}$ in component form and in terms of magnitude and direction.

  • Guidance: differentiate to obtain velocity components:

  • $v_x(t) = \frac{dx}{dt} = 0.5 t,$

  • $v_y(t) = \frac{dy}{dt} = 1.0 + 0.075 t^2,$

  • Instantaneous speed: $|\mathbf{v}(t)| = \sqrt{v<em>x(t)^2 + v</em>y(t)^2}.$

  • At $t=2\text{s}$: $vx(2) = 1.0\ \text{m/s}$, $vy(2) = 1.3\ \text{m/s}$; speed ≈ $\sqrt{1.0^2+1.3^2} \approx 1.64\ \text{m/s}$; direction $\theta = \tan^{-1}(vy/vx) \approx 52.5^{\circ}$ above the +x axis.

• Example: Remote-controlled car – velocity as a function of time (problem sketch from slides)

  • Velocity components given by

  • $\mathbf{v}(t) = \bigl[5.00 - (0.0180) t^2\bigr]\hat{\mathbf{i}} + \bigl[2.00 + 0.550 t\bigr]\hat{\mathbf{j}},$

  • where units are in m/s for components and t in seconds.

  • Tasks:

  • (a) Find the $x$- and $y$-components of velocity as functions of time: $v<em>x(t) = 5.00 - 0.0180 t^2,$ $v</em>y(t) = 2.00 + 0.550 t.$

  • (b) Magnitude and direction of the velocity at $t = 8.00\ \text{s}$.

  • (c) Magnitude and direction of the acceleration at $t = 8.00\ \text{s}$ (acceleration is the time derivative of velocity).

Projectile-motion problems (typical problems and formulas)

• Problem 1 (from slides): A ball is kicked horizontally at speed $8.0\ \text{m/s}$ from a cliff of height $80\ \text{m}$. How far from the base of the cliff will it land?

  • Approach: horizontal motion with constant velocity; vertical motion under gravity; time to fall from height $h$ is determined by $y(t) = h - \tfrac{1}{2} g t^2$, solve for $t$, then horizontal range $x = v_{0x} t$.

• Problem 2: A shell is fired from a cliff at speed $800\ \text{m/s}$ at an angle $30^{\circ}$ below the horizontal, to reach a point $150\ \text{m}$ below the launch height. Find time to hit, etc. (setup involves solving vertical motion with a final $y$-position of $-150\ \text{m}$ and then using horizontal range if needed.)

• Problem 3: A baseball is thrown with horizontal component $v_{0x} = 25\ \text{m/s}$ and takes $3.00\ \text{s}$ to return to its initial height. Determine horizontal range, initial vertical velocity component, and initial launch angle.

• Problem 4: A bullet fired at $60^{\circ}$ with $v_0 = 200\ \text{m/s}$. How long is the bullet in the air? What is the maximum height reached?

Exact projectile equations and their derivations (summary)

• Initial conditions: $x<em>0=0,\quad y</em>0=0,$ $v<em>{0x} = v</em>0 \cos \theta<em>0,\quad v</em>{0y} = v<em>0 \sin \theta</em>0.$

• Horizontal motion (no acceleration):

  • $x(t) = x<em>0 + v</em>{0x} t,$

  • $v<em>x(t) = v</em>{0x}.$

• Vertical motion (constant downward acceleration):

  • $y(t) = y<em>0 + v</em>{0y} t - \tfrac{1}{2} g t^2,$

  • $v<em>y(t) = v</em>{0y} - g t.$

• Trajectory (eliminate t between x and y):

  • $y(x) = x \tan\theta<em>0 - \frac{g x^2}{2 v</em>0^2 \cos^2\theta_0}.$

• Range and height results (for level ground):

  • $R = \frac{v<em>0^2 \sin(2\theta</em>0)}{g},$

  • $h<em>{\max} = \frac{v</em>0^2 \sin^2\theta_0}{2 g},$

  • $T = \frac{2 v0 \sin\theta0}{g}$, the total time of flight, when landing at the same height as launch.

• Peak time and apex: $t<em>{\text{peak}} = \frac{v</em>0 \sin\theta_0}{g}.$

• Vertical velocities at launch and return to same height have equal magnitude and opposite sign: $vy(T) = -v0 \sin\theta_0$ for symmetric flight (no air resistance).

• Geometry of vectors in 2D remains: independent components, trajectory as a parabola, and the right-triangle decomposition of vectors.

Hints for solving 2D motion problems (from slides)

• Define a coordinate system with axes and origin.

• List known quantities and determine components: $v{0x}$, $v{0y}$, $ax$, $ay$, etc.

• Use the fact that time is the same for both horizontal and vertical motions.

• Separate the problem into horizontal and vertical subproblems:

  • Horizontal: $x = x0 + v{0x} t + \frac{1}{2} a_x t^2$.

  • Vertical: $y = y0 + v{0y} t + \frac{1}{2} a_y t^2$.

• Determine which equations apply (constant acceleration in each axis).

• Solve for time(s) when a given condition occurs (e.g., $y=0$ for range, $y=h$ for height).

Essential equations (compact reference)

• Position vectors and components:

  • $\mathbf{r}(t) = x(t)\hat{\mathbf{i}} + y(t)\hat{\mathbf{j}}.$

• Displacement: $\Delta \mathbf{r} = \Delta x\hat{\mathbf{i}} + \Delta y\hat{\mathbf{j}}.$

• Velocity and acceleration components:

  • $v<em>x(t) = \frac{dx}{dt}, \quad v</em>y(t) = \frac{dy}{dt},$

  • $a<em>x(t) = \frac{d^2x}{dt^2}, \quad a</em>y(t) = \frac{d^2y}{dt^2}.$

• Trajectory: $y(x) = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}.$

• Horizontal range, maximum height, and time of flight:

  • $R = \frac{v<em>0^2 \sin(2\theta)}{g}, \quad T = \frac{2 v</em>0 \sin\theta}{g}, \quad h<em>{\max} = \frac{v</em>0^2 \sin^2\theta}{2 g}.$

Quick reference: notational recap

• 2D position: $\mathbf{r}(t) = x(t)\hat{\mathbf{i}} + y(t)\hat{\mathbf{j}}.$

• 2D velocity: $\mathbf{v}(t) = v<em>x(t)\hat{\mathbf{i}} + v</em>y(t)\hat{\mathbf{j}}.$

• 2D acceleration: $\mathbf{a}(t) = a<em>x(t)\hat{\mathbf{i}} + a</em>y(t)\hat{\mathbf{j}}.$