Physics › Mechanics › Projectile motion

Projectile motion

A projectile is two motions at once: constant velocity horizontally and constant acceleration vertically, with neither affecting the other. Time is the only quantity common to both, so every projectile calculation is solved by finding the time in one direction and using it in the other.

Year 12AQA 3.4.1.4 CIE 2.1.9 OCR A 3.1.3(a), 3.1.3(b)EDEXCEL 9, 15

Builds on Motion graphs and the SUVAT equations.

The maths behind it: Projectiles on InkMaths.

IN THIS TOPIC

Explain the independence of horizontal and vertical motion in a uniform gravitational field.
Solve projectile problems by treating the two directions separately, linked only by time.
Describe qualitatively how air resistance changes a projectile's trajectory.

COMMON MISCONCEPTION

Something must keep pushing a projectile forwards.

Two motions, one clock

Once a projectile leaves your hand, ignoring air resistance, exactly one force acts on it. Its weight, straight down. Nothing pushes it forward, and nothing has to, because motion continues perfectly well without a force. Out of that comes the central fact of the topic, that the horizontal and vertical motions are independent.

FIG. 1One ball dropped, one launched sideways, released at the same instant. The amber dashed line joins them and never tilts: their vertical motions are the same motion. On the launched ball, the horizontal velocity arrow never changes while the vertical one grows steadily from nothing, which is free fall at g. Faded strobes freeze the story frame by frame.

Drop one ball and launch another horizontally at the same instant, and they hit the ground together. The launched ball's sideways speed does nothing to its fall. Horizontally it moves at constant velocity; vertically it accelerates at g like anything else in free fall.

Solving the two problems

Every projectile question is the same procedure. Resolve the initial velocity into components. Treat the vertical motion as constant acceleration, a = g downward, using the SUVAT equations. Treat the horizontal motion as constant speed, distance = speed × time. The single shared quantity is time, so almost every problem finds t from one direction and spends it in the other.

FIG. 2Velocity components along the flight. The horizontal component never changes; the vertical one shrinks to zero at the top and grows again coming down.

The velocity picture follows from that. The horizontal component is identical at every point of the flight, while the vertical one passes through zero at the apex. At the top of the arc the projectile is still moving horizontally, so “velocity at the highest point” has a non-zero answer, and students who write zero throw the mark away.

WORKED EXAMPLE

The full two-column method

A ball is struck from level ground at 25 m s⁻¹, 30° above the horizontal. Find its time of flight and its range. Take g = 9.81 m s⁻².

Draw it before you calculate anything, then resolve the launch velocity. Across, 25 cos 30° = 21.7 m s⁻¹. Up, 25 sin 30° = 12.5 m s⁻¹. Two columns, one clock.

The flight time lives in the vertical column. Landing at launch height means the vertical displacement is zero, so s = ut + ½at² with s = 0 gives t = 2 × 12.5/9.81 = 2.55 s.

Now spend that time in the horizontal column, where there is no acceleration. Range = 21.7 × 2.55 = 55 m to two significant figures.

Is it reasonable? A hard strike carries a ball tens of metres, not hundreds, and a hang time near two and a half seconds is what you see watching one. Every input was SI, so the answer arrived in metres without being asked.

GUIDED PRACTICE

How high does the same ball go?

Same launch: 25 m s⁻¹ at 30°, so the vertical component is 12.5 m s⁻¹ and at the top the vertical velocity is zero. Choose the vertical SUVAT equation with no t in it, solve for the height, then compare with the working.

Show the working

With v = 0, u = 12.5 and a = −9.81, the equation without t is $v^{2} = u^{2} + 2as$ , so 0 = 12.5² − 2 × 9.81 × s.

s = 156.25/19.62 = 8.0 m. Worth checking against a rough climb, about 1.3 s at an average vertical speed near 6 m s⁻¹, which lands in the same region.

INDEPENDENT PRACTICE

A ramp and a landing

A stunt rider leaves a ramp at 18 m s⁻¹, 40° above the horizontal, and lands at the same height. How far away is the landing? Work it through completely.

Show the working

Resolve first. Across, 18 cos 40° = 13.8 m s⁻¹. Up, 18 sin 40° = 11.6 m s⁻¹.

In the vertical column, s = 0 gives t = 2 × 11.6/9.81 = 2.36 s.

In the horizontal column, range = 13.8 × 2.36 = 33 m to two significant figures. The two columns never mixed. Only t crossed between them.

What air resistance does

A qualitative account is all that is asked for. Air resistance opposes motion and grows with speed, so it steals horizontal speed throughout the flight and fights the vertical motion both ways, up and down.

FIG. 3The same launch with and without air resistance. The real path peaks lower, lands shorter and comes down more steeply than it went up.

The resulting trajectory is lower and shorter than the ideal parabola, and it loses its symmetry. The descent comes down steeper than the ascent went up, because by then the horizontal speed has been eaten away. That same argument caps a vehicle's top speed. Resistive forces climb with speed until they meet the driving force, and where they meet is the maximum.

ASSESSMENT FOCUS

Start by resolving the launch velocity and ruling two columns, horizontal and vertical. Mixing components inside one equation is the defining error of this topic.
The horizontal direction has no acceleration. Never put g in it, never use SUVAT there, and remember distance is just speed times time.
At the apex the vertical velocity is zero and the speed is not, because the horizontal component survives. Both halves of that get asked.
For air resistance, three phrases score. Lower maximum height, shorter range, descent steeper than ascent.

CHECK YOURSELF

A ball rolls off a table 1.25 m high at 4.0 m s⁻¹. How long is it in the air, and how far from the table does it land?

Show a hint

The fall time comes from the vertical motion alone. The sideways speed has nothing to do with it.

Show the answer

Vertically, $s = \tfrac{1}{2}gt^{2}$ with s = 1.25 m gives $t^{2} = 2 \times 1.25 / 9.81 = 0.255$ , so t = 0.50 s.

Horizontally, a constant 4.0 m s⁻¹ for 0.50 s gives a range of 4.0 × 0.50 = 2.0 m.

Notice the order of the work. The vertical problem produced the time, and the horizontal problem spent it.

Across, constant velocity.

Down, constant acceleration.

Time is the only thing the two columns share.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the projectile motion questions page.

4 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

Explain the independence of horizontal and vertical motion in a uniform gravitational field.
Solve projectile problems by treating the two directions separately, linked only by time.
Describe qualitatively how air resistance changes a projectile's trajectory.

Open the full revision checklist to track your progress across the whole unit.

Projectile motion

Two motions, one clock

Solving the two problems

What air resistance does

Question workbook

Question workbook

Answer book