Mechanics

1D Kinematics

Start

Learn position, velocity, and acceleration through graph slopes and a stop-in-zones challenge.

2D Kinematics

Here

Extend kinematics into vector components, projectile motion, and two-dimensional acceleration.

Collisions

Finish

Compare elastic and inelastic collisions while keeping momentum and energy bookkeeping straight.

2D Kinematics

In one dimension, position, velocity, and acceleration each need one signed number, or scalar. In two dimensions, each quantity becomes a vector with horizontal and vertical components:

\begin{aligned} \vec{r} &= \langle x, y \rangle,\\ \vec{v} &= \langle v_x, v_y \rangle,\\ \vec{a} &= \langle a_x, a_y \rangle. \end{aligned}

The definitions are the same, but we now track motion along two perpendicular axes.

Component Motion

The useful trick is to split the vector equation into component equations. If acceleration is constant, then

\vec{v}(t) = \vec{v}_i + \vec{a}t,

and

\vec{r}(t) = \vec{r}_i + \vec{v}_it + \frac{1}{2}\vec{a}t^2.

Written as separate horizontal and vertical equations, this becomes

\begin{aligned} v_x(t) &= v_{ix} + a_xt,\\ v_y(t) &= v_{iy} + a_yt, \end{aligned}

\begin{aligned} x(t) &= x_i + v_{ix}t + \frac{1}{2}a_xt^2,\\ y(t) &= y_i + v_{iy}t + \frac{1}{2}a_yt^2. \end{aligned}

Note that these are all functions of a common variable $t$ . This shared time is what lets us combine the separate one-dimensional calculations into a two-dimensional path.

Launch Components

Projectile problems often describe an initial velocity using a launch speed $v_0$ and angle $\theta$ . It’s common to use a $0$ subscript in place of $i$ , when $t=0$ is implicitly the initial time. The horizontal and vertical pieces are

v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta.

Launch Decomposition

Drag the launch vector or use the sliders to see how speed and angle set the horizontal and vertical velocity components.

Projectile Motion

A simple projectile has no horizontal acceleration and a constant downward acceleration from gravity:

\vec{a} = \langle 0, -g \rangle, \qquad g \approx 9.8\,\text{m/s}^2.

If the projectile starts at the origin, the component equations become

\begin{aligned} x(t) &= v_{0x}t,\\ y(t) &= v_{0y}t - \frac{1}{2}gt^2, \end{aligned}

\begin{aligned} v_x(t) &= v_{0x},\\ v_y(t) &= v_{0y} - gt. \end{aligned}

The horizontal component of velocity stays constant while the vertical component changes steadily due to gravity. A projectile effectively moves sideways at a steady rate while its height rises, pauses, and then falls.

Why is the projectile path a parabola?

Start with the horizontal equation:

x = v_{0x}t.

If $v_{0x} \neq 0$ , then

t = \frac{x}{v_{0x}}.

Substitute that time into the vertical equation:

y = v_{0y}\left(\frac{x}{v_{0x}}\right) - \frac{1}{2}g\left(\frac{x}{v_{0x}}\right)^2.

With $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$ , this can be written as

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

That is a quadratic function of $x$ , so the trajectory is a parabola.

Projectile Launcher

Aim, launch, and compare the predicted range with the actual path. Add drag to see where the ideal equations stop being exact.

Acceleration in 2D

For small intervals of time $\Delta t$ ,

\begin{aligned} \Delta \vec{v} &\approx \vec{a}\Delta t \end{aligned}

This equation can be interpreted as “acceleration changes the velocity vector”. If the acceleration and velocity vectors are aligned, the velocity gets larger in time: the object speeds up. If acceleration points opposite the velocity, the velocity gets smaller: the object slows down. What happens when acceleration is perpendicular to velocity?

2D Acceleration Sandbox

Apply a constant acceleration to an object in two dimensions.

Problem Solving

Soccer Ball Components

Problem

A soccer ball is kicked at $20\,\text{m/s}$ at an angle of $30^\circ$ above the horizontal. Find its initial velocity components.

Given

$v_0 = 20\,\text{m/s}$
$\theta = 30^\circ$

Use

v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta.

Solve

v_{0x} = (20)\cos 30^\circ \approx 17.3\,\text{m/s},

v_{0y} = (20)\sin 30^\circ = 10.0\,\text{m/s}.

Check

The launch angle is closer to horizontal than vertical, so the horizontal component should be larger than the vertical component. It is.

Time of Flight and Range

Problem

Using the same soccer ball, find the time to maximum height, total flight time, and horizontal range. Assume it lands at the same height it was kicked from and ignore air resistance.

Given

$v_{0x} \approx 17.3\,\text{m/s}$
$v_{0y} = 10.0\,\text{m/s}$
$g = 9.8\,\text{m/s}^2$

Solve

At maximum height, $v_y = 0$ :

0 = v_{0y} - gt \qquad \Rightarrow \qquad t = \frac{v_{0y}}{g} = \frac{10.0}{9.8} \approx 1.02\,\text{s}.

For a level landing, the trip down mirrors the trip up, so

T \approx 2(1.02) = 2.04\,\text{s}.

The horizontal range is

R = v_{0x}T \approx (17.3)(2.04) \approx 35.3\,\text{m}.

Check

A two-second flight at a little over $17\,\text{m/s}$ horizontally should cover a bit more than $34\,\text{m}$ , so $35.3\,\text{m}$ is reasonable.

2D Kinematics Checkpoint

Question 1 of 3

In ideal projectile motion with no air resistance, what happens to the horizontal velocity?

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