The kinematics equations: which one to use, worked line by line on a projectile

There are five kinematics equations, and each one leaves out exactly one of the five quantities: displacement, initial velocity, final velocity, acceleration, time. So the rule for choosing is a sentence: list what you know and what you want, find the one quantity you neither know nor want, and use the equation that does not contain it. Below, the five equations in a table, the rule applied, and one projectile worked line by line with units carried and the wrong line most students write.

The five kinematics equations and what each leaves out

The five equations describe one object moving in a straight line with a constant acceleration. That condition matters: a car speeding up smoothly, a ball falling, a puck sliding to a stop on ice all qualify. A car in stop-and-go traffic does not. Within that condition there are five quantities, and each equation connects four of them.

Equation	The quantity it leaves out
v = v₀ + at	displacement Δd
Δd = v₀t + ½at²	final velocity v
v² = v₀² + 2aΔd	time t
Δd = ½(v₀ + v)t	acceleration a
Δd = vt − ½at²	initial velocity v₀

Some courses write the same five with other letters, s for displacement and u for initial velocity, and call them the suvat equations. Same five equations, same rule. Some write them as "kinematic equations" without the s. Same thing again.

Throughout this page the acceleration of a falling object is g = 9.8 m/s², the value the SPH4U course uses. Every answer is given to three significant figures unless the data carry fewer.

The rule for choosing one: the missing quantity

Four steps, and they are a habit, not a trick. Do them in this order every time until the order is automatic.

1. Write the givens with units and signs. Every number the question hands you, with its unit, and with a sign that says which way it points. Pick a positive direction before you write the first sign.
2. Write the unknown. The one quantity the question asks for, as a symbol with a question mark.
3. Name the quantity that appears in neither list. There are five quantities. You know some, you want one. Usually exactly one is left over.
4. Pick the equation without that quantity. Read it off the table.

A small case with no projectile in it. A car brakes from 20.0 m/s to rest over 50.0 m. What is its acceleration?

Givens: v₀ = 20.0 m/s, v = 0 m/s, Δd = 50.0 m, forward positive. Unknown: a. The quantity in neither list is t, so the equation is v² = v₀² + 2aΔd.

• 0² = 20.0² + 2 × a × 50.0 (the equation with the numbers in)
• 0 = 400 + 100a (squaring and multiplying, in m²/s² and m)
• a = −400 ÷ 100 = −4.0 m/s² (solving; the minus means the acceleration points backwards, against the motion, which is what braking is)

Two significant figures, because 20.0 and 50.0 carry three but the question's answer only needs to say −4.0 against a "forward positive" convention; write −4.00 m/s² if your teacher wants three. If the question had asked for the stopping time instead, the quantity in neither list would have been Δd, and the equation v = v₀ + at: 0 = 20.0 − 4.0t gives t = 5.0 s.

Projectile motion is two one-dimensional problems sharing a time

A projectile is anything thrown or launched and then left alone to gravity. Its motion looks like one curved path. It is better treated as two straight-line motions happening at once.

Horizontally, nothing pushes on the projectile once it is in flight, so a = 0 and the horizontal speed never changes. The only equation you need on that axis is Δx = vₓt.

Vertically, gravity acts the whole time, so the acceleration is g = 9.8 m/s² downward. You choose once whether down is positive (then a = +9.8 m/s²) or up is positive (then a = −9.8 m/s²), and you keep that choice for every vertical line in the problem. The five equations apply on this axis.

The two axes share exactly one quantity: the time. The projectile is in the air for one length of time, and that time is the same whether you are watching it move sideways or watching it fall. So the standard method is to solve the vertical axis for t and then hand that t to the horizontal axis. "The horizontal speed never changes" is the whole trick, because it means the horizontal axis has nothing to solve until the time arrives.

Worked example: a ball rolls off a 20.0 m cliff at 15.0 m/s

A ball rolls horizontally off the edge of a cliff 20.0 m high at 15.0 m/s. How long does it fall, how far from the base of the cliff does it land, and how fast is it moving when it lands?

Givens: h = 20.0 m (the vertical displacement), vₓ = 15.0 m/s (constant), v₀ᵧ = 0 m/s (a horizontal launch has no vertical speed to begin with), g = 9.8 m/s². Down is positive for the vertical axis, so a = +9.8 m/s². Three significant figures.

Line 1, the time to fall. Vertical axis. Known: Δy = 20.0 m, v₀ᵧ = 0, a = 9.8 m/s². Wanted: t. In neither list: vᵧ. The equation without vᵧ is Δd = v₀t + ½at².

• 20.0 m = 0 × t + ½ × 9.8 m/s² × t² (the equation with the numbers in)
• 20.0 m = 4.9 m/s² × t² (multiplying ½ × 9.8)
• t² = 20.0 ÷ 4.9 = 4.08 s² (dividing both sides; the unit is m ÷ m/s² = s²)
• t = √4.08 = 2.02 s (taking the root; the positive root, since the ball lands after it leaves)

Line 2, the range. Horizontal axis, a = 0, so Δx = vₓt.

• Δx = 15.0 m/s × 2.02 s = 30.3 m (multiplying; m/s × s = m)

The ball lands 30.3 m from the base of the cliff. The time came from the vertical axis and was handed across; nothing else did.

Line 3, the vertical speed at the ground. Vertical axis again. Known: v₀ᵧ = 0, a = 9.8 m/s², t = 2.02 s. Wanted: vᵧ. In neither list: Δd. The equation without Δd is v = v₀ + at.

• vᵧ = 0 + 9.8 m/s² × 2.02 s = 19.8 m/s (multiplying; m/s² × s = m/s), downward

Line 4, the speed on landing. The two components are at right angles, so the speed is the hypotenuse.

• speed = √(15.0² + 19.8²) = √(225 + 392) = √617 = 24.8 m/s (Pythagoras; the numbers inside are in m²/s²)

The energy route gives the same number without the time: √(vₓ² + 2gh) = √(15.0² + 2 × 9.8 × 20.0) = √(225 + 392) = √617 = 24.8 m/s. The work–energy theorem is why that shortcut is allowed; use it as a check, not as a replacement, until the course reaches it.

Line 5, the direction. The angle below the horizontal has the vertical component opposite and the horizontal component adjacent.

• tan θ = 19.8 ÷ 15.0 = 1.32 (opposite over adjacent, both in m/s, so the ratio has no unit)
• θ = tan⁻¹(1.32) = 52.9° below the horizontal

Five lines, each one chosen by naming the missing quantity first. Write that name in the margin beside each line when you practise; it is the thing a marker is checking for when the method marks are awarded.

The three wrong lines a marker sees

Hestenes, Wells and Swackhamer (1992) built the Force Concept Inventory because students arrive in mechanics with a stable set of beliefs about motion that the course has to displace, not just add to. Projectile problems show three of them as wrong lines, and each has a symptom you can spot in your own work.

1. The 15.0 m/s goes into the vertical equation as v₀ᵧ. The horizontal speed is not a vertical speed. A ball rolled off a flat edge has v₀ᵧ = 0. The symptom: a fall time that depends on how fast the ball was rolling, which it does not. Drop a ball and roll one off the same table at the same moment and they hit the floor together.
2. A sign mixed inside one line. Down taken as positive for g and up as positive for a velocity in the same equation. The symptom: a negative t², a time that is "undefined", or a square root of a negative number on the calculator. Choose the positive direction once, write it at the top of the page, and give every vertical quantity its sign against that one choice.
3. v = d/t used for the fall. That equation is the a = 0 case, Δx = vₓt. The vertical axis has a = 9.8 m/s², so the average speed during the fall is not the final speed and not zero either. The symptom: a time that is about twice too short or twice too long, depending on which speed was used.

A fourth, briefer: forgetting that the time is shared, and computing the range with a time invented for the horizontal axis. The horizontal axis cannot produce a time on its own; it has one equation and two unknowns until the vertical axis supplies t.

One worked example is not enough on its own. Sweller and Cooper (1985) found that students who studied worked examples and then solved a matching problem learned the procedure faster than students who only solved problems; why a worked example teaches is that the example shows the decision at each line, which is exactly what "name the missing quantity" is. The next section is the matching problem.

A sibling to try cold

Cover everything above. A stone rolls off a 45.0 m cliff at 12.0 m/s. Find the time to fall, the range, and the speed on landing. Before you write any equation, write the five givens with units and signs, write the unknown for the first line, and name the quantity that appears in neither list. Only then pick the equation. Redo a worked example cold, then a sibling is the pattern for every explainer on this site: the answer below is for checking, not for reading first.

Check

• Time: Δy = ½gt² → t² = 2 × 45.0 ÷ 9.8 = 9.18 s² → t = 3.03 s
• Range: Δx = 12.0 m/s × 3.03 s = 36.4 m
• Vertical speed: vᵧ = 9.8 m/s² × 3.03 s = 29.7 m/s
• Speed: √(12.0² + 29.7²) = √(144 + 882) = √1,026 = 32.0 m/s

If your time was right and your range wrong, look at line 2 of the worked example again. If your time was wrong, look at which quantity you put in as v₀ᵧ.

Where this sits in the course

This is idea 1 of 25 in SPH4U, Two-dimensional kinematics and projectile motion, the first idea of strand B. It assumes the Grade 11 one-dimensional work: the five equations above on one axis, and vectors broken into components. If the table in the first section was new to you, that is the thing to go back for before the projectile.

The idea after it is idea 2, Newton's laws; free-body diagrams; net force, which keeps the same habit, two axes and one shared quantity, but the shared quantity becomes the acceleration and the axes carry forces instead of velocities. Newton's laws and free-body diagrams, the idea after this one is written the same way as this page. The SPH4U course map shows all 25 ideas in teaching order, and every idea of the three science courses has a page like this one.

The kinematics equations on one screen

• Five equations, five quantities. Each equation leaves out one of Δd, v₀, v, a, t. Constant acceleration only.
• The choosing rule. Givens with units and signs; the unknown; the quantity in neither list; the equation without it.
• Projectiles. Horizontal: a = 0, Δx = vₓt. Vertical: a = 9.8 m/s² with a sign you chose once. The time is the only thing the axes share.
• The three wrong lines. The horizontal speed used as v₀ᵧ; a sign mixed inside one line; v = d/t on the axis that accelerates.
• The method. Derive the equation choice from the missing quantity every time rather than memorising which equation goes with which problem type; how to study physics is built on that habit.