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:
Ax = A \, \cos\theta, \qquad Ay = A \ sin\theta,
A = \sqrt{Ax^2 + Ay^2}, \qquad \theta = \tan^{-1}\left( rac{Ay}{Ax}\right).
In 2D, vectors are written as
\mathbf{A} = Ax \hat{\mathbf{i}} + Ay \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}(t2) - \mathbf{r}(t1) = \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) = x0 + v{0x} t + \frac{1}{2} a_x t^2,
Vertical motion: y(t) = y0 + v{0y} t + \frac{1}{2} a_y t^2,
with the corresponding velocities vx(t) = v{0x} + ax t, \quad vy(t) = v{0y} + ay 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\theta0 - \frac{g x^2}{2 v0^2 \cos^2\theta_0}.
Initial conditions for projectile problems:
x0 = 0,\quad y0 = 0,
v{0x} = v0 \cos\theta0, \quad v{0y} = v0 \sin\theta0,
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 = x0 + v{0x} t, (if $a_x=0$)
Vertical motion: y = y0 + v{0y} t - frac{1}{2} g t^2,
Vertical velocity: vy(t) = v{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{\max} = \frac{v0^2 \sin^2\theta}{2 g}.
Peak time: t{\text{peak}} = \frac{v0 \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: vx(t) = v0 \cos\theta, (for idealized constant $v{0x}$) and vy(t) = v_0 \sin\theta - g t.
The speed magnitude: |\mathbf{v}(t)| = \sqrt{vx(t)^2 + vy(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{0x} = v0 \cos 25^{\circ} \approx 9.06\ \text{cm/s},
v{0y} = v0 \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{vx(t)^2 + vy(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: vx(t) = 5.00 - 0.0180 t^2, vy(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: x0=0,\quad y0=0, v{0x} = v0 \cos \theta0,\quad v{0y} = v0 \sin \theta0.
Horizontal motion (no acceleration):
x(t) = x0 + v{0x} t,
vx(t) = v{0x}.
Vertical motion (constant downward acceleration):
y(t) = y0 + v{0y} t - \tfrac{1}{2} g t^2,
vy(t) = v{0y} - g t.
Trajectory (eliminate t between x and y):
y(x) = x \tan\theta0 - \frac{g x^2}{2 v0^2 \cos^2\theta_0}.
Range and height results (for level ground):
R = \frac{v0^2 \sin(2\theta0)}{g},
h{\max} = \frac{v0^2 \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{\text{peak}} = \frac{v0 \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:
vx(t) = \frac{dx}{dt}, \quad vy(t) = \frac{dy}{dt},
ax(t) = \frac{d^2x}{dt^2}, \quad ay(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{v0^2 \sin(2\theta)}{g}, \quad T = \frac{2 v0 \sin\theta}{g}, \quad h{\max} = \frac{v0^2 \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) = vx(t)\hat{\mathbf{i}} + vy(t)\hat{\mathbf{j}}.
2D acceleration: \mathbf{a}(t) = ax(t)\hat{\mathbf{i}} + ay(t)\hat{\mathbf{j}}.