Unit 1 · Topic 1.5

1.5 Vectors and Motion in Two Dimensions

Any vector can be split into perpendicular x and y components with trigonometry, and that turns two-dimensional motion into two separate one-dimensional problems that share the same clock. Projectile motion is the classic case: constant velocity sideways, constant acceleration g downward.

Key terms

vector components
resultant
trigonometry (sin, cos, tan)
projectile motion
horizontal and vertical motion
time of flight

Breaking vectors into components

A vector of magnitude A at angle θ above the +x axis has components A_x = A cos θ and A_y = A sin θ. Going the other way, the magnitude is A = √(A_x² + A_y²) and the angle satisfies tan θ = A_y/A_x.

These come from right-triangle trigonometry: the vector is the hypotenuse, and the components are the legs. If the angle is measured from a different line, such as from vertical, sine and cosine swap, so always check which side is adjacent to the angle.

The equation sheet lists sine, cosine and tangent for common angles. Note 37° and 53°, which come from a 3-4-5 triangle: sin 37° ≈ 3/5 and cos 37° ≈ 4/5.

To add vectors in two dimensions, add the x components, add the y components, then rebuild the magnitude and direction if you need them.

Two-dimensional motion as two 1D problems

Motion in a plane can be analyzed with the one-dimensional kinematic equations if you split it into x and y parts. The two directions are independent: what happens horizontally doesn't change what happens vertically. The only thing they share is time, and time is usually the link that solves the problem.

Projectile motion

A projectile is any object moving through the air with only gravity acting on it. With up as +y:

The horizontal acceleration is zero, so v_x stays constant the whole flight. The vertical acceleration is a_y = −g, so v_y changes by g every second.

At the top of its path, v_y = 0 but v_x isn't zero (unless it was thrown straight up). So the projectile's speed is smallest at the top, not zero.

A ball launched horizontally and a ball dropped from the same height at the same moment hit level ground at the same time, because both start with v_y = 0. The launched ball just covers horizontal distance while it falls.

Graph	Horizontal (x)	Vertical (y)
position vs. time	straight line	downward-opening parabola
velocity vs. time	flat line at v_x	straight line, slope −g
acceleration vs. time	zero	flat line at −g

Launches from a height

If a projectile is launched at an angle from a cliff or a rooftop, it doesn't land at its launch height, so the up-and-down symmetry no longer holds. Find the time from the vertical equation y = y₀ + v₀y t − ½gt², setting y to the landing height. That's usually a quadratic, and you keep the positive root. Then use that time in the horizontal equation.

Range and launch angle

On level ground with no air resistance, launch angles that add to 90° (such as 30° and 60°) give the same horizontal range, and 45° gives the longest range for a given speed. The steeper launch stays in the air longer and goes higher. There's no range formula on the equation sheet; you can derive it from the kinematic equations when you need it.

Air resistance is ignored unless a problem says otherwise. If it's included, you only reason about it qualitatively: the projectile goes less high and less far, and its path is no longer a symmetric parabola.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Ball rolling off a table
A ball rolls off a 1.25 m high table at 3.0 m/s. Use g = 9.8 m/s². How long is it in the air, and how far from the table's edge does it land?
Show the solutionHide the solution
1. Step 1: Split the motion. Horizontal: v_x = 3.0 m/s, a_x = 0. Vertical (down as positive here): v₀y = 0, a_y = 9.8 m/s², Δy = 1.25 m.
2. Step 2: Vertical gives the time: Δy = ½gt², so t = √(2Δy/g) = √(2 × 1.25 ÷ 9.8) ≈ 0.505 s.
3. Step 3: Horizontal uses that same time: Δx = v_x t = 3.0 × 0.505 ≈ 1.52 m.
Answer: About 0.51 s in the air; it lands about 1.5 m from the table
Example 2Calculator allowed
Launch at an angle on level ground
A ball is kicked from level ground at 20 m/s, 37° above horizontal. Use g = 10 m/s², sin 37° ≈ 0.60 and cos 37° ≈ 0.80, and ignore air resistance. Find (a) the time to reach the top, (b) the maximum height, (c) the total time in the air and (d) the horizontal range.
Show the solutionHide the solution
1. Step 1: Components: v_x = 20 cos 37° ≈ 16 m/s; v₀y = 20 sin 37° ≈ 12 m/s.
2. Step 2: (a) At the top v_y = 0: 0 = 12 − 10t, so t = 1.2 s.
3. Step 3: (b) v_y² = v₀y² − 2gh: 0 = 144 − 20h, so h = 7.2 m.
4. Step 4: (c) It lands at its launch height, so the flight is symmetric: total time = 2 × 1.2 = 2.4 s.
5. Step 5: (d) Range = v_x × total time = 16 × 2.4 = 38.4 m. Note that only the vertical motion set the time; the horizontal speed just carries the ball along.
Answer: (a) 1.2 s; (b) 7.2 m; (c) 2.4 s; (d) about 38 m
Example 3Calculator allowed
Comparing two launch angles (classic trap)
Two balls are launched from level ground at the same speed v₀, one at 30° and one at 60°. A student says the 60° ball must land farther away because it's in the air longer. Is she right?
Show the solutionHide the solution
1. Step 1: Derive the range. Time in the air: T = 2v₀ sin θ / g. Range: R = (v₀ cos θ)(T) = 2v₀² sin θ cos θ / g.
2. Step 2: For 30°: sin 30° cos 30° = (0.5)(0.866) ≈ 0.433. For 60°: sin 60° cos 60° = (0.866)(0.5) ≈ 0.433. The products are identical, so the ranges are equal.
3. Step 3: The 60° ball is in the air longer (sin 60° > sin 30°), but its horizontal velocity is smaller (cos 60° < cos 30°). The two effects cancel exactly.
4. Step 4: With v₀ = 20 m/s and g = 10 m/s², both land 34.6 m away; the 30° ball is in the air 2.0 s and the 60° ball about 3.5 s.
Answer: No. Both land the same distance away; the 60° ball has a longer flight time but a smaller horizontal velocity.

Common mistakes

Saying a projectile's velocity is zero at the top. Only v_y is zero; v_x is unchanged, so the speed there equals v_x.
Using the full launch speed in a one-direction equation. Use v₀ cos θ horizontally and v₀ sin θ vertically (when θ is measured from horizontal).
Giving the horizontal motion an acceleration. With no air resistance, a_x = 0 for the whole flight.
Thinking a faster horizontal launch makes an object fall more slowly. Fall time depends only on the vertical motion.

On the exam

Free-response projectile questions usually ask you to separate components, find the time from the vertical motion and then use it horizontally. Label each equation x or y so readers can follow you.
You may be asked to sketch v_x–t and v_y–t graphs for a projectile, or to compare two launches. Flat line for v_x, straight line with slope −g for v_y.
Lab questions can ask how to predict where a launched ball lands: measure the launch height and speed, then use the time to fall.

Connected topics

Videos

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