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Free fall at $g$, projectiles as independent horizontal and vertical motions, angled launches, and when air resistance breaks the model.
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By the end of this lesson you will be able to analyze falling objects and projectiles and judge when air resistance matters.
From the last lessons you know the constant-acceleration equations and how to read motion graphs. From Physics with numbers you know that gravity pulls everything down. This lesson applies the equations to falling objects and to projectiles, whose motion splits into two simpler parts, and asks when the model stops working.
| Term | What it means |
|---|---|
| Free fall | Motion under gravity alone, with acceleration $g = 9.8$ m/s² downward. |
| Projectile | An object moving under gravity alone after being launched. |
| Components | The horizontal and vertical parts of a velocity: $v_0\cos\theta$ and $v_0\sin\theta$. |
| Time of flight | How long a projectile is in the air, set by its vertical motion. |
| Range | The horizontal distance a projectile covers. |
| Air resistance | The drag force of the air, neglected in the simple model. |
Near Earth's surface, ignoring air resistance, every object in free fall accelerates downward at $g = 9.8$ m/s², regardless of its mass. A projectile moves in two directions at once, and the two motions do not affect each other:
The time of flight comes from the vertical motion alone; the horizontal distance is then $v_xt$. The path is a parabola. For a launch at speed $v_0$ and angle $\theta$, $v_x = v_0\cos\theta$ and $v_y = v_0\sin\theta$.
Another way: picture
Picture two balls leaving a table at the same moment: one simply dropped, one knocked sideways. A strobe photograph shows them level with each other at every flash, falling side by side, even as the second one moves across. Its sideways motion does nothing to its fall, and the fall does nothing to its sideways motion.
Another way: steps
The figure shows a ball launched horizontally from a cliff, drawn at equal time intervals. The gray marks along the top show its horizontal position: equally spaced, because its horizontal velocity never changes. The marks down the side show its fall: the gaps grow as one, three, five and seven units, the pattern of steady acceleration from rest.
Put the two together and the path curves into a parabola. The time to reach the ground is exactly the time a dropped ball would take from the same height; a faster launch simply carries the ball farther in that time.
Galileo argued, and experiments confirm, that in the absence of air all objects fall with the same acceleration. A bowling ball and a feather dropped in a vacuum chamber hit the floor together. On the Moon in 1971, astronaut David Scott of Apollo 15 dropped a hammer and a falcon feather, and they landed at the same instant.
The reason, which Newton's laws explain, is that a heavier object feels a larger gravitational force but also has more inertia, and the two exactly cancel. In air, the feather flutters because air resistance is large compared with its tiny weight.
A ball thrown upward slows by $9.8$ m/s every second, stops at the top, and falls back, speeding up by $9.8$ m/s each second. The motion is symmetric: it takes as long to rise as to fall, and it returns to the launch height at the speed it left with, now pointing down.
The maximum height is $v_0^2/(2g)$: double the launch speed and the ball rises four times as high. A ball tossed up at $10$ m/s rises about $5$ m and is in the air about two seconds, which is why juggling higher throws takes much more time between catches.
A ball kicked at an angle has both horizontal and vertical velocity. Split the launch velocity with trigonometry: $v_x = v_0\cos\theta$ and $v_y = v_0\sin\theta$. The vertical part sets how long it stays up, $2v_y/g$ on level ground; the horizontal part sets how far it goes in that time.
Without air resistance, a launch at $45°$ gives the greatest range for a given speed, and launches at complementary angles, like $30°$ and $60°$, land at the same spot. The steeper one stays up longer and goes higher, the shallower one gets there sooner.
Checking an answer. The fall time must not depend on the horizontal speed. The vertical velocity must be zero at the top. The landing speed on level ground must equal the launch speed.
The two directions are independent because gravity acts only vertically; with air resistance neglected, nothing pushes horizontally. Each component obeys its own constant-acceleration equations, with $a_x = 0$ and $a_y = -g$.
The value $9.8$ m/s² holds near Earth's surface; it varies slightly with latitude and altitude, from about $9.78$ at the equator to $9.83$ at the poles, and falls off far above the surface. For projectiles that stay low and slow, the model is excellent.
Air resistance grows with speed, roughly as the square of the speed for most objects. For a baseball hit at $45$ m/s, drag cuts the range by nearly half compared with the vacuum prediction, and the best launch angle drops to about $35°$. For a skydiver, drag eventually balances weight, and the fall reaches a terminal speed of about $55$ m/s.
Light objects with large surfaces, like leaves, paper and feathers, never follow the free-fall model at all. The model works best for dense, compact objects moving slowly over short distances: a dropped stone, a tossed ball, water from a hose.
Physics students measure g by dropping objects past light gates, by timing pendulums, or by filming a falling ball and analyzing the video frame by frame. Precision instruments called absolute gravimeters drop a mirror in a vacuum and track it with a laser, measuring g to a billionth of its value.
The National Geodetic Survey maps small variations in g across the United States. Dense rock beneath the surface pulls slightly harder, so gravity maps help find mineral deposits and oil, and they help define elevations precisely.
Every thrown, kicked or hit ball is a projectile. A basketball free throw arcs from about $2$ m up to a rim $3.05$ m high, $4.2$ m away; players who shoot with a higher arc give the ball a better angle to drop through the hoop. A long jumper is a projectile too, launching at about $20°$ because they cannot generate much vertical speed while sprinting.
Coaches use projectile models, with corrections for air resistance and spin, to advise athletes. Statcast tracking in Major League Baseball measures every batted ball's launch speed and angle, and home runs cluster around launch angles of $25°$ to $35°$.
A stream of water from a hose follows a parabola, each drop a separate projectile. Fountain designers use this to shape displays: the Bellagio fountains in Las Vegas launch water up to $140$ m, timing jets so that the arcs meet the music.
Firefighters use the same physics: tilting a hose nozzle up increases the reach, with the best angle a little under $45°$ because air resistance slows the spray. A hose stream reaching $30$ m needs a nozzle speed of about $20$ m/s, which the pumps on a fire engine must supply against friction in the hoses.
On level ground, a projectile's flight is symmetric about its highest point. It rises for half the flight time and falls for the other half, and at each height it passes with the same speed going up as coming down.
That symmetry makes hang time a direct measure of height: a punt that hangs for $4.5$ s rose for $2.25$ s and reached about $25$ m above the kicker's foot. It also means a projectile's range can be found from the time up alone, doubling it for the whole trip, a shortcut that saves solving a quadratic equation.
NFL coaches and scouts time punts with stopwatches from the kick until the catch. A hang time of $4.5$ s is considered excellent: it gives the coverage team time to run downfield and surround the returner before the ball arrives.
Treating the punt as a projectile, it rises for half the hang time, $2.25$ s, and falls for the other half, so its peak is $\tfrac{1}{2}g(2.25)^2$, about $25$ m above the kicker's foot. A $5.0$ s punt reaches about $31$ m. Real punts are slowed by air resistance and spin, so the peak is somewhat lower, but the symmetry of the flight makes hang time the simplest measure of height a coach can take from the sideline.
On August 2, 1971, at the end of the Apollo 15 moonwalk, astronaut David Scott held a geological hammer in one hand and a falcon feather in the other and let them go together in front of the television camera. With no air on the Moon, they fell side by side and hit the ground at the same moment.
The Moon's gravity is about a sixth of Earth's, $1.6$ m/s², so from a height of $1.6$ m they took about $1.4$ s to fall, long enough to watch clearly. The demonstration, which Scott called a tribute to Galileo, is still shown in physics classes, and NASA's video of it is among the most watched clips from the Apollo program.
It is natural to think a ball launched fast horizontally stays up longer, as if its speed held it up. The vertical and horizontal motions are independent: a ball fired horizontally and a ball dropped from the same height land at the same moment. The fast one simply lands farther away.
A related error is to think heavier objects fall faster. Without air resistance, all objects fall with the same acceleration. In air, dense compact objects come close to that ideal, while light spread-out ones fall much more slowly.
A marble rolls off a $0.80$ m high table at $2.5$ m/s. Find the fall time.
$t = \sqrt{\dfrac{2 \times 0.80}{9.8}} = 0.404\ \text{s}$
Vertical motion alone.
Find the horizontal distance.
$x = 2.5 \times 0.404 = 1.01\ \text{m}$
Constant horizontal speed.
Find the vertical speed at landing.
$v_y = 9.8 \times 0.404 = 3.96\ \text{m/s}$
Downward.
Find the landing speed.
$v = \sqrt{2.5^2 + 3.96^2} = 4.68\ \text{m/s}$
Combine the parts.
Double the launch speed. Find the new fall time.
$t = 0.404\ \text{s}$
Unchanged: height alone decides.
A baseball is popped straight up at $24.5$ m/s. Find the time to the top.
$t = \dfrac{24.5}{9.8} = 2.5\ \text{s}$
$v_y$ falls to zero.
Find the maximum height.
$h = \dfrac{24.5^2}{2 \times 9.8} = 30.6\ \text{m}$
Time-free equation.
Find the total time back to the bat's height.
$2 \times 2.5 = 5.0\ \text{s}$
Symmetric flight.
Find its velocity at $4.0$ s.
$v = 24.5 - 9.8 \times 4.0 = -14.7\ \text{m/s}$
Falling.
Find its height at $4.0$ s.
$y = 24.5 \times 4.0 - 4.9 \times 16 = 19.6\ \text{m}$
On the way down.
Compare with its height at $1.0$ s.
$y = 24.5 - 4.9 = 19.6\ \text{m}$
Same height, rising: symmetry.
A soccer ball is kicked at $22$ m/s, $35°$ above level ground. Find the horizontal part.
$v_x = 22\cos 35° = 18.0\ \text{m/s}$
Cosine.
Find the vertical part.
$v_y = 22\sin 35° = 12.6\ \text{m/s}$
Sine.
Find the time in the air.
$t = \dfrac{2 \times 12.6}{9.8} = 2.57\ \text{s}$
Up and down.
Find the range.
$R = 18.0 \times 2.57 = 46.3\ \text{m}$
Horizontal speed times time.
Find the peak height.
$h = \dfrac{12.6^2}{2 \times 9.8} = 8.1\ \text{m}$
From the vertical part.
Predict the effect of air resistance.
$\text{shorter range, lower peak}$
Real balls fall short.
Write the drop equation.
$h = \tfrac{1}{2}gt^2$
From rest.
Substitute the values.
$h = \tfrac{1}{2} \times 9.8 \times 3.0^2$
Square the time.
Evaluate the distance.
A ball rolls off a horizontal ledge $1.25$ m high at $6$ m/s. Ignoring air resistance and taking $g = 9.8$ m/s², how far from the base of the ledge does it land, in m?
Complete the worked solution: a soccer ball is kicked from level ground at $25$ m/s, $53.13°$ above the horizontal. Ignoring air resistance with $g = 9.8$ m/s², find the horizontal and vertical parts of its launch velocity in m/s, and its time in the air in s.
Find the horizontal part.
$v_x = v_0\cos\theta =$ x
Adjacent to the angle.
Find the vertical part.
$v_y = v_0\sin\theta =$ y
Opposite the angle.
Find the time in the air.
$t = \dfrac{2v_y}{g} =$ t
Up and back to the same level.
Find the range from the parts.
$R = v_x\,t$
Constant horizontal speed.
Match each part of projectile motion to the rule it follows, ignoring air resistance.
| stays constant throughout the flight | 9.8 m/s² downward the whole time | zero | decided by the vertical motion alone | |
|---|---|---|---|---|
| horizontal velocity | ||||
| vertical acceleration | ||||
| vertical velocity at the highest point | ||||
| time in the air |
A stone is dropped from rest from a bridge $80$ m above a river. Ignoring air resistance with $g = 9.8$ m/s², fill in the fall time in s, the speed on hitting the water in m/s, and the average speed during the fall in m/s.
| value | |
|---|---|
| fall time (s) | |
| landing speed (m/s) | |
| average speed (m/s) |
A ball is thrown straight up from ground level at $12$ m/s. Ignoring air resistance with $g = 9.8$ m/s² and up as positive, write its height in meters as a function of time $t$ in seconds.
Answer:
A ball is thrown straight up at $20$ m/s. Ignoring air resistance with $g = 9.8$ m/s², how high above the launch point does it rise, in m?
Answer: m
NFL scouts time punts by their hang time, from the kick until the catch at the same height. A punt hangs for $4.2$ s. Ignoring air resistance with $g = 9.8$ m/s², how high did it rise above the kick, in m?
Answer: m
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A ball is thrown straight up from ground level at $10$ m/s. Ignoring air resistance with $g = 9.8$ m/s² and up as positive, write its height in meters as a function of time $t$ in seconds.
Answer:
You can analyze projectiles. Explain to someone why a ball knocked off a table lands at the same moment as one dropped beside it.
22. Your turn: a stone is dropped from rest and falls for $3.0$ s. How far does it fall?, step 3
$h = 44.1\ \text{m}$
Meters.