Projectiles - Physics for Pilot Assessments

A projectile is anything thrown, fired or dropped that then moves under gravity alone. The single idea that makes these problems tractable is that the horizontal and vertical parts of the motion are completely independent — neither affects the other — so a problem that looks two-dimensional is really two simple one-dimensional problems side by side.

Two motions, no interaction

Once a projectile has left your hand, and ignoring air resistance, the only force acting on it is its weight, which points straight down. That has an immediate consequence: there is no horizontal force at all, so by Newton's first law the horizontal velocity never changes. Meanwhile the vertical motion is exactly free fall, accelerating downwards at g regardless of what is happening sideways.

The result that seems wrong

Fire a bullet horizontally and drop another from the same height at the same instant, and they hit the ground together. The fired one travels hundreds of metres while the dropped one falls at your feet, but they touch down simultaneously, because the vertical motion does not know or care about the horizontal. Nothing about going sideways fast keeps an object up.

Both balls at equal time intervals — one dropped, one fired sideways. Each dashed line joins them at the same instant, and they are always at the same height.

This is what makes projectile problems manageable. The two motions share exactly one quantity: the time. So the usual route through a question is to use the vertical motion to find how long the flight lasts, then use that time in the horizontal motion to find how far it went.

Both halves of that route come with a small formula, and neither is new. Falling from rest, the speed gained is just the acceleration multiplied by the time — the definition of acceleration rearranged. The distance fallen follows from an averaging trick: the vertical speed climbs steadily from zero to g × t, so its average over the fall is half of g × t, and multiplying that average by the time gives half of g times the time squared.

Speed gained falling from rest

v = g t

$v$: vertical speed after falling for time t, in metres per second (m/s)
$g$: the free-fall acceleration — 10 m/s² in every question here
$t$: time spent falling, in seconds (s)

Distance fallen from rest

s = \tfrac{1}{2} g t^2

$s$: vertical distance fallen, in metres (m)
$g$: the free-fall acceleration, 10 m/s²
$t$: time spent falling, in seconds (s)

Horizontal distance covered

\text{range} = v_h \times t

$v_h$: the horizontal speed, in metres per second (m/s) — constant for the whole flight
$t$: the total time in the air, in seconds (s) — found from the vertical motion alone

Check them against each other with a two-second fall: $v = 10 \times 2 = 20$ m/s, and $s = \tfrac{1}{2} \times 10 \times 2^2 = \tfrac{1}{2} \times 10 \times 4 = 20$ m — the same 20 metres the average-speed argument gives, because it is the same argument written compactly. The examples below use whichever form is quicker.

Why the path is a curve

The shape follows from the two motions combined. Horizontally the projectile covers equal distances in equal times, because that velocity is constant. Vertically it covers increasing distances in equal times, because it is speeding up. Put them together and the path droops more and more steeply as it goes — the curve known as a parabola. A ball thrown upwards at an angle traces the same shape rising and falling, symmetrical about its highest point, and at that highest point its vertical velocity is momentarily zero while its horizontal velocity is exactly what it always was.

Putting it to work

Launching at an angle

Firing at an angle rather than horizontally changes nothing about the method. The launch velocity now has both a horizontal and a vertical part, so you separate them at the start: the horizontal part stays constant for the whole flight, exactly as before, while the vertical part starts upwards, is slowed by gravity, reaches zero at the top, then grows downwards. The projectile rises, curves over and falls, and the shape it traces is the same parabola.

Splitting the launch velocity is the same triangle you met when two perpendicular forces were combined, run backwards. There the 3 and the 4 at right angles gave a resultant of 5; here a launch of 25 m/s comes apart into 15 m/s across and 20 m/s up — the same 3-4-5, scaled by five. Questions choose their numbers from that family precisely so the split needs no calculator.

A 25 m/s launch resolved into its two parts: 15 m/s horizontally and 20 m/s vertically. The horizontal part is the same at every point of the flight, which is why only it survives at the top of the arc and why the landing velocity is the launch velocity mirrored.

The symmetry of that arc is worth noticing because it saves work. Launched from level ground, the projectile takes exactly as long to come down as it took to go up, and it lands with the same speed it left with, just directed downwards instead of upwards. So if you know it took two seconds to reach the top, the whole flight lasted four. And at the very top the vertical velocity is momentarily zero while the horizontal velocity is untouched — so the projectile is at its slowest there, but it is certainly not stationary.

Example 3 — a launch at an angle, start to finish

A ball leaves level ground at 25 m/s, angled so that the launch splits into 15 m/s horizontally and 20 m/s vertically. Taking g = 10, how high does it climb, how long is it in the air, and how far away does it land?

Separate the two parts and keep them separate. Horizontal: 15 m/s, and it stays 15 m/s for the entire flight. Vertical: 20 m/s upwards, and this is the only part gravity acts on.
Time to the top comes from the vertical part alone. It loses 10 m/s of upward speed every second, so 20 m/s is used up after $\frac{20}{10} = 2$ s.
Height climbed: the vertical speed falls evenly from 20 m/s to zero, so the average over the climb is $\frac{20 + 0}{2} = 10$ m/s. Over 2 s that gives 10 × 2 = 20 m.
Total time in the air: from level ground the descent mirrors the climb, so the flight lasts 2 × 2 = 4 s.
Range needs the horizontal part and the total time — nothing else. distance = speed × time = 15 × 4 = 60 m.
Two checks worth making. At the top the ball still moves at 15 m/s horizontally, so it is at its slowest there but not stationary. And it lands at 25 m/s, the launch speed again, now pointing downwards.

One further consequence of the independence deserves stating, because it is the basis of a whole family of questions. If two projectiles are launched from the same height at the same instant with different horizontal speeds, they land at the same time but at different distances — and the ratio of those distances is exactly the ratio of their speeds. Double the launch speed and you double the range, because the flight time was identical and range is simply speed multiplied by that time. This is a rare case of a clean proportional relationship in mechanics, and it holds only because the vertical motion is untouched by the horizontal.

What air resistance does to all this

Everything above ignores air resistance, which for anything light or fast is a genuine simplification rather than a small correction. Drag acts against the direction of travel at every instant, which means it is the one force in this lesson that has a horizontal part — and that alone is enough to break the clean picture, because the horizontal velocity no longer stays at its launch value. Questions in this course say "ignore air resistance" or describe a dense, compact object precisely so that the clean version applies, but it is worth knowing which direction reality departs in, and why.

The useful way to think about drag is to split it the same way the motion was split. Horizontally there is now a backwards force where previously there was none at all, so the horizontal velocity decays throughout the flight and the projectile covers a little less ground in each successive second. Vertically, drag changes sides halfway through: on the way up it points downwards, adding to gravity, so upward speed is lost faster than 10 m/s each second and the climb is cut short both in time and in height. On the way down it points upwards, opposing gravity, so downward speed builds more slowly than it otherwise would, and instead of growing without limit it creeps towards the terminal velocity met in the free-fall lesson.

Put those together and the symmetry that made the ideal case so convenient is gone. Because the ascent is hurried and the descent is held back, the descent takes longer than the climb — not the same time. Because the horizontal velocity has been decaying the whole way, less ground is covered coming down than going up. Those two facts pull in opposite directions and both are worth holding onto: the highest point arrives before the halfway mark in time, yet sits past the halfway mark in distance. The final approach to the ground is therefore steeper than the launch, the object lands slower than it left rather than at its launch speed, and the range falls short of the ideal figure — the lighter and the larger the object, the further short it falls.

The same launch drawn twice. Both paths leave at the same angle and speed, because drag has not acted yet — from then on the real path peaks lower, leans its summit past the midpoint of its own range, comes down more steeply than it went up, and lands short.

Example 4 — reading the difference between the two paths

The launch above gave an ideal range of 60 m. A light plastic ball, launched identically in still air, lands about 45 m away and visibly steeper than it left. Account for the missing 15 m and for the steeper landing — and say what would have to be true for the ideal answer to have been right.

Start with the horizontal, because that is where the ideal calculation gets its range. The ideal answer used 15 m/s for all 4 s. With drag the ball starts at 15 m/s and slows continuously, so its average horizontal speed over the flight is below 15 m/s, and less ground is covered. That is the bulk of the missing 15 m.
The flight time changes too. Drag adds to gravity going up, so the climb ends sooner and lower than the ideal 2 s and 20 m; coming down, drag opposes gravity, so that shortened fall is slower and takes longer than the climb did. The tidy "time up = time down" is the first casualty of leaving air resistance in, and the ideal 4 s no longer applies either.
Now the steeper landing. Horizontal speed at the end is well below 15 m/s, while vertical speed has grown towards its terminal value. A path's steepness is the ratio of vertical to horizontal speed, and drag has shrunk the bottom of that ratio while the top kept growing — so the ball arrives at a steeper angle than the 3-4-5 launch.
It also lands slower than 25 m/s. Gravity gave back on the way down exactly what it took on the way up, but drag removed energy continuously in both directions and returned none of it.
For the ideal answer to be right, drag would have to be negligible beside the object's weight — dense, compact and not too fast. A steel ball of the same size would land far closer to 60 m than the plastic one, which is exactly why questions specify the object, and why "ignore air resistance" is stated rather than assumed.

Why this matters beyond the test

The independence of the two motions is not a trick for answering questions; it is why anything thrown can be aimed at all. A thrown ball, a hose jet, a long jump and a fired shell all follow the same arc, and in each case the person aiming is unconsciously solving the same two-part problem: how long will it be in the air, and how far will it travel sideways in that time. Understanding that the two parts do not interfere is what makes the arc predictable rather than mysterious, and it is the reason a single simple rule covers cases that look nothing like each other.

Why it matters in aviation

The habit this lesson builds — treat the vertical and the horizontal as two separate problems that share only the clock — is exactly how a descent is planned, and it is one of the first pieces of mental arithmetic a pilot uses routinely. An aircraft that has to lose 9,000 feet does not think about a curve through the sky. The vertical question is a rate: at 1,500 feet per minute, 9,000 feet takes six minutes. The horizontal question then uses that same six minutes: at 300 knots groundspeed, six minutes covers 30 nautical miles, so the descent has to start at least 30 miles out. Time is the only quantity passed between the two halves, which is precisely the structure of using the fall to find the flight duration and then the horizontal speed to find the range.

The independence also decides what happens to anything that leaves an aircraft in flight, and this is where the intuition most often fails. A container released from a cargo aircraft does not drop straight down and does not fly backwards: it keeps the aeroplane's forward velocity at the moment of release and merely begins to accelerate downwards on top of it. Ignoring drag it would stay directly beneath the aircraft the whole way; in practice drag slows its forward travel so it falls progressively behind, but it still touches down far ahead of the point of release — hundreds of metres ahead, from any normal drop height. That is why an air-drop release point is computed rather than judged by eye, and the same reasoning covers ice shed from an airframe and a jettisoned load. It also disposes of the standard exam claim that a faster aircraft's dropped load stays airborne longer: it does not. The height alone sets the fall time, and the extra speed buys nothing but extra distance.

What the test asks

The headline trap is that a faster horizontal launch stays airborne longer — it does not, because the time is decided entirely by the vertical motion and the height. Expect an item comparing a dropped and a fired object, where the correct answer is that they land together. And watch for the top of an arc, where a distractor claims the projectile is momentarily stationary: its vertical velocity is zero there, but the horizontal velocity is untouched, so it is still moving.