Motion in two dimensions and projectiles

Position, velocity and acceleration as vectors in a plane, and why a thrown ball follows a parabola.

Follows MIT 8.01SC lesson 3MIT 8.01SC Lesson 3: 2D Kinematics – Position, Velocity, and Acceleration (3.1–3.5)On MIT OpenCourseWare: the lecture and its readings

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Vectors (8.01)
Position and velocity in one dimension (8.01)
Acceleration in one dimension (8.01)

On this page · 6 sections

Position, velocity and acceleration as vectors
Projectile motion
The path is a parabola
Aiming at a falling target
Try it
Summary

So far everything moved along a line. Real motion happens in space, but the step from one dimension to two (or three) is smaller than it looks. With vectors, a plane is just two lines handled at the same time.

Position, velocity and acceleration as vectors#

Choose an origin and Cartesian axes, with unit vectors $\hat i$ and $\hat j$ . The object's location is the position vector, and its velocity and acceleration are its time derivatives, taken one component at a time:

Kinematics in two dimensions

\vec r(t) = x(t)\,\hat i + y(t)\,\hat j, \qquad \vec v(t) = \frac{dx}{dt}\,\hat i + \frac{dy}{dt}\,\hat j, \qquad \vec a(t) = \frac{dv_x}{dt}\,\hat i + \frac{dv_y}{dt}\,\hat j

In words

vector r (t) equals x(t) i hat plus y(t) j hat , vector v (t) equals dx over dt i hat plus dy over dt j hat , vector a (t) equals dv sub x over dt i hat plus dv sub y over dt j hat

Source: 8.01SC Course Notes, Chapter 5: Two Dimensional Kinematics, Sections 5.1–5.2.1 (all equations are numbered 5.1.x), Eqs. (5.1.1)–(5.1.3)

This works because the unit vectors of a Cartesian system never change direction, so only the components change with time. Each component behaves exactly like the one-dimensional motion of the previous pages: $v_x = dx/dt$ , $a_x = dv_x/dt$ , and the same for $y$ .

The velocity vector always points along the path, tangent to it, in the direction of motion. The acceleration vector need not. It points in the direction the velocity is changing, which can be sideways or even backwards.

Projectile motion#

A projectile is anything thrown, kicked or launched that then moves only under gravity. Near Earth's surface, and ignoring air resistance, gravity gives a constant acceleration of size $g \approx 9.8\ \text{m/s}^2$ pointing straight down. With $+y$ up, the force is $\vec F_g = -mg\,\hat j$ and Newton's second law gives

Accelerations of a projectile

a_x = 0, \qquad a_y = -g

In words

a sub x equals 0, a sub y equals minus g

Source: 8.01SC Course Notes, Chapter 5: Two Dimensional Kinematics, Sections 5.1–5.2.1 (all equations are numbered 5.1.x), Sec. 5.2, Eqs. (5.1.14)–(5.1.15)

The mass drops out: a heavy ball and a light ball follow the same path. This is Galileo's law of free fall, now in two dimensions.

Launch at speed $v_0$ and angle $\theta_0$ above the horizontal. The initial velocity components are

Initial velocity components

v_{x,0} = v_0\cos\theta_0, \qquad v_{y,0} = v_0\sin\theta_0

In words

v sub x,0 equals v sub 0 cos theta sub 0 , v sub y,0 equals v sub 0 sine theta sub 0

Source: 8.01SC Course Notes, Chapter 5: Two Dimensional Kinematics, Sections 5.1–5.2.1 (all equations are numbered 5.1.x), Eqs. (5.1.5)–(5.1.6)

Now apply the constant-acceleration results from the previous lesson to each direction separately:

Projectile position and velocity

\vec r(t) = \big(x_0 + v_{x,0}\,t\big)\,\hat i + \big(y_0 + v_{y,0}\,t - \tfrac12 g t^2\big)\,\hat j, \qquad \vec v(t) = v_{x,0}\,\hat i + \big(v_{y,0} - g t\big)\,\hat j

In words

vector r (t) equals (x sub 0 plus v sub x,0 t ) i hat plus (y sub 0 plus v sub y,0 t minus one half g t squared ) j hat , vector v (t) equals v sub x,0 i hat plus (v sub y,0 minus g t ) j hat

Source: 8.01SC Course Notes, Chapter 5: Two Dimensional Kinematics, Sections 5.1–5.2.1 (all equations are numbered 5.1.x), Eqs. (5.1.16)–(5.1.17) with a_y = −g

The path is a parabola#

Solve $x(t) = x_0 + v_{x,0}t$ for $t$ and substitute it into $y(t)$ . With the origin at the launch point ( $x_0 = y_0 = 0$ ) the result is the orbit equation:

Trajectory (orbit equation)

y(x) = \frac{v_{y,0}}{v_{x,0}}\,x - \frac{g}{2\,v_{x,0}^2}\,x^2

In words

y(x) equals v sub y,0 over v sub x,0 x minus g over 2 v sub x,0 squared x squared

Source: 8.01SC Course Notes, Chapter 5: Two Dimensional Kinematics, Sections 5.1–5.2.1 (all equations are numbered 5.1.x), Sec. 5.2.1, Eq. (5.1.33)

That is a downward-opening parabola. Its slope at any point, $dy/dx = v_y/v_x$ , gives the direction of the velocity there, but not the speed. Change the launch speed, angle and height below.

Loading graph…

v_0

\theta_0

h

Trajectories computed from the orbit equation (g = 9.8 m/s², launched from height h). On level ground, 45° gives the longest range.

Worked example

A ball rolled off a cliff

A ball rolls off the edge of a 20 m high cliff with a horizontal speed of $8.0\ \text{m/s}$ . Ignoring air resistance, how long is it in the air, how far from the foot of the cliff does it land, and how fast is it moving when it lands?

Set up coordinates

Origin at the foot of the cliff, $+x$ outward, $+y$ up. Then $x_0 = 0$ , $y_0 = 20$ m, $v_{x,0} = 8.0\ \text{m/s}$ , $v_{y,0} = 0$ .

Use the vertical motion to find the time

The ball lands when $y = 0$ : $0 = 20 - \tfrac12(9.8)t^2$ , so $t = \sqrt{40/9.8} = 2.02$ s. The horizontal speed played no part: a ball simply dropped from the cliff lands at the same moment.

Use the horizontal motion to find the distance

$x = v_{x,0}\,t = 8.0 \times 2.02 = 16.2$ m from the foot of the cliff.

Find the landing velocity

$v_x = 8.0\ \text{m/s}$ (unchanged) and $v_y = -9.8 \times 2.02 = -19.8\ \text{m/s}$ . The speed is $\sqrt{8.0^2 + 19.8^2} = 21.4\ \text{m/s}$ , at $\tan^{-1}(19.8/8.0) = 68^\circ$ below the horizontal.

Worked example

Range on level ground

A ball is kicked from the ground at $15\ \text{m/s}$ , $35^\circ$ above the horizontal, and lands at the same height. Find its time of flight, maximum height and range.

Components of the initial velocity

$v_{x,0} = 15\cos 35^\circ = 12.3\ \text{m/s}$ , $v_{y,0} = 15\sin 35^\circ = 8.60\ \text{m/s}$ .

Time of flight

It lands when $y = v_{y,0}t - \tfrac12 g t^2 = 0$ , so $t = 2v_{y,0}/g = 2(8.60)/9.8 = 1.76$ s.

Maximum height

At the top $v_y = 0$ , halfway through the flight at $t = 0.88$ s. Then $y_\text{max} = v_{y,0}^2/(2g) = 8.60^2/19.6 = 3.78$ m.

Range

$x = v_{x,0}\,t = 12.3 \times 1.76 = 21.6$ m. In general the range is $2v_{x,0}v_{y,0}/g = v_0^2\sin(2\theta_0)/g$ , which is largest at $\theta_0 = 45^\circ$ , as the plot above shows.

Aiming at a falling target#

MIT's lesson ends with a classic demonstration: a target is released at the instant a ball is launched straight at it. The ball hits whatever its speed, as long as it reaches the target before either hits the ground. The equations above explain why. Both the ball and the target fall the same distance $\tfrac12 g t^2$ in the same time, so gravity shifts both down by exactly the same amount. Relative to the target, the ball still travels along the straight line it was aimed along.

Try it#

Fire the cannon at different angles and speeds, and turn air resistance on and off. Without air resistance the paths are the parabolas above; with it, the range is shorter and the path is no longer symmetric.

Projectile Motion. Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY-NC 4.0 (https://phet.colorado.edu). Open full screen

Summary#

In a plane, $\vec r$ , $\vec v$ and $\vec a$ are vectors; each Cartesian component obeys the one-dimensional rules.
A projectile has $a_x = 0$ and $a_y = -g$ : constant horizontal velocity, free fall vertically, sharing one clock.
Eliminating time gives a parabolic path $y(x)$ ; its slope gives the direction of motion.
On level ground the range is $v_0^2 \sin(2\theta_0)/g$ , largest at $45^\circ$ (without air resistance).

Symbols on this page

Show the 9 symbols used here

$\theta$ theta: An angle. Say it: “theta” (sounds like “thay-ta or thee-ta”).
$\circ$ degree: Degrees of angle (as in 45°). Say it: “degrees”.
$\approx$ approximately equal: Is approximately equal to. Say it: “is approximately”.
$\times$ times: Times (multiplication). Between two vectors: the cross product. Say it: “times”.
$\sin$ sine: Sine of an angle. Say it: “sine”.
$\cos$ cosine: Cosine of an angle. Say it: “cos”.
$\tan$ tangent: Tangent of an angle, sin/cos. Say it: “tan”.
$\vec v$ vector arrow: A vector: a quantity with a size and a direction. Say it: “vector”.
$\hat\imath$ hat: A unit vector: a vector of length 1 that only gives a direction (î, r̂). Say it: “hat”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

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Check your understanding

A ball is thrown at an angle above the horizontal. Ignoring air resistance, what happens to the horizontal component of its velocity during the flight?
At the highest point of a projectile's path, the velocity is…
Ignoring air resistance, two balls leave a table edge at the same moment: one is dropped, one is pushed horizontally at $3\ \text{m/s}$ . Which lands first?
What shape is the path $y(x)$ of a projectile without air resistance?

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