Sample: a physics handout, before and after

This is the same handout in HTML. Here the equations are written in MathML, so screen readers that support math can read them aloud and let you move through them term by term.

What assistive technology gets from each file
Part of the file	Before	After
Title and language	Not set	Document title shown, language set to English
Headings and bookmarks	None	A heading for every section, plus bookmarks
Equations	Drawings with no text behind them, so a screen reader skips them	Tagged as formulas, each with spoken alt text
Figure 1, the graph	Skipped	Alt text that describes the curve and its key values
Table 1	A run of numbers with no headers	Column and row headers linked to every cell
Lists	Plain lines of text	Tagged lists, with numbers on the worked steps

Projectile Motion: the setup

A ball is launched from level ground with speed v₀ at an angle θ above the horizontal. If we ignore air resistance, the only force on the ball after launch is gravity, which pulls it down with acceleration g. The horizontal and vertical motions can then be treated separately.

Position at time t

The horizontal velocity never changes, so the horizontal position grows steadily with time:

x (t) = v_{0} \cos θ t

The vertical velocity starts at v₀ sin θ and falls at the rate g, which gives:

y (t) = v_{0} \sin θ t - \frac{1}{2} g t^{2}

Three useful results

Setting y(t) = 0 gives the time of flight T. Putting T into x(t) gives the range R, and the top of the path gives the maximum height H.

Time of flight $T = \frac{2 v_{0} \sin θ}{g}$
Range $R = \frac{v_{0}^{2} \sin 2 θ}{g}$
Maximum height $H = \frac{v_{0}^{2} \sin^{2} θ}{2 g}$

Graph of height against horizontal distance. A symmetric curve rises from the origin to a peak of 5.10 meters at a distance of about 17.7 meters, then falls back to the ground at 35.3 meters. — Figure 1. Path of a ball launched at 20 m/s and 30°. It reaches a maximum height of 5.10 m and lands 35.3 m away.

Worked example

A ball is launched at v₀ = 20 m/s and θ = 30°. Take g = 9.8 m/s². Find the time of flight, the range and the maximum height.

Time of flight
$T = \frac{2 \times 20 \times \sin 30 °}{9.8} = 2.04 s$
Range
$R = \frac{20^{2} \times \sin 60 °}{9.8} = 35.3 m$
Maximum height
$H = \frac{20^{2} \times \sin^{2} 30 °}{2 \times 9.8} = 5.10 m$

Changing the launch angle

Table 1 keeps v₀ = 20 m/s and changes only the angle.

Table 1. Flight time, maximum height and range for v₀ = 20 m/s and g = 9.8 m/s²
Launch angle	Time of flight (s)	Maximum height (m)	Range (m)
15°	1.06	1.37	20.4
30°	2.04	5.10	35.3
45°	2.89	10.2	40.8
60°	3.53	15.3	35.3
75°	3.94	19.0	20.4

Two things stand out. The range is largest at 45°, and angles that add up to 90°, such as 30° and 60°, give the same range. The higher angle keeps the ball in the air longer and sends it higher.

Try it yourself

At what launch angles would the ball land 20.4 m away?
If v₀ doubled, how would the range change?

Answers: 15° or 75°. The range would become four times larger.

A physics handout, before and after

What changed

The handout as a web page