Unit 1 · Topic 1.5

1.5 Motion in Two or Three Dimensions

Motion in two or three dimensions splits into perpendicular components that move independently, each following its own one-dimensional kinematics. Projectile motion is the main example: constant horizontal velocity and constant downward acceleration g. When position is given as a vector function of time, you differentiate each component.

Key terms

projectile motion
components
position vector
independence of perpendicular motion
trajectory

Independence of perpendicular motion

Horizontal and vertical motion don't affect each other. Drop one ball and fire another horizontally from the same height at the same moment, and (ignoring air resistance) they hit level ground together. The fired ball covers horizontal distance, but its vertical motion is identical to the dropped ball's.

So you solve a 2D problem as two 1D problems that share one thing: time. Time is the link between the x-equations and the y-equations.

Projectile motion

A projectile moves under gravity alone, ignoring air resistance. Take +x horizontal and +y up. For a launch at speed v₀ and angle θ above the horizontal:

Quantity	Horizontal (x)	Vertical (y)
acceleration	0	$-g$
initial velocity	$v_0\cos\theta$	$v_0\sin\theta$
velocity at time t	$v_0\cos\theta$ (constant)	$v_0\sin\theta - gt$
position at time t	$x_0 + (v_0\cos\theta)t$	$y_0 + (v_0\sin\theta)t - \frac{1}{2}gt^2$

Features of the path

The path (trajectory) is a parabola. At the highest point the vertical velocity is zero, but the horizontal velocity is not, so the speed there is $v_0\cos\theta$ , the slowest point of the flight. The acceleration is g downward the whole time, including at the top.

On level ground, the trip up and the trip down take the same time, and the projectile lands at the same speed it was launched with. Complementary angles (like 30° and 60°) give the same range, and 45° gives the maximum range on level ground. Those are level-ground results; when launch and landing heights differ, work it out from the equations.

A typical strategy: find time from whichever direction gives it most easily (often vertical), then use that time in the other direction.

A plan for projectile problems

Most projectile questions fall to the same routine. Writing it out on free response also shows the reader your reasoning.

Pick axes and an origin, usually +y up with the origin at the launch point, and write down g with the correct sign.
Split the launch velocity into $v_{0x}$ and $v_{0y}$ .
Write the x-equation and the y-equation separately. List what you know and what you need in each.
Find the time from the direction where you have enough information, usually y. For a landing height below the launch point, solve the quadratic and keep the positive root.
Use that time in the other direction, then combine components with the Pythagorean theorem if a speed or angle is asked.

Position, velocity and acceleration as vector functions

In Physics C you may get the position vector as a function of time, like $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$ . Differentiate each component separately:

$\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}, \qquad \vec{a} = \frac{d\vec{v}}{dt}$

The speed is the magnitude of $\vec{v}$ . The velocity vector is always tangent to the path. Motion in three dimensions works the same way with a $\hat{k}$ term, but on the exam it's described rather than calculated.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Rolling off a table
A ball rolls off a horizontal table 1.25 m high at 3.0 m/s. Ignore air resistance and use g = 9.8 m/s². How far from the table's edge does it land, and how fast is it moving just before it lands?
Show the solutionHide the solution
1. Step 1: Vertical: it starts with zero vertical velocity and falls 1.25 m. $1.25 = \frac{1}{2}(9.8)t^2$ , so $t = \sqrt{2(1.25)/9.8} \approx 0.505$ s.
2. Step 2: Horizontal: constant 3.0 m/s, so x = (3.0)(0.505) ≈ 1.52 m.
3. Step 3: Final velocity components: $v_x = 3.0$ m/s and $v_y = -(9.8)(0.505) \approx -4.95$ m/s.
4. Step 4: Speed: $\sqrt{3.0^2 + 4.95^2} \approx 5.8$ m/s, at about 59° below the horizontal.
Answer: It lands about 1.5 m from the edge, moving about 5.8 m/s.
Example 2Calculator allowed
Launch at an angle on level ground
A ball is kicked from level ground at 25 m/s, 37° above the horizontal. Ignore air resistance and use g = 9.8 m/s². Find the time of flight, the range and the maximum height.
Show the solutionHide the solution
1. Step 1: Components: $v_{0x} = 25\cos 37^\circ \approx 20.0$ m/s and $v_{0y} = 25\sin 37^\circ \approx 15.0$ m/s (15.05 before rounding; keep the unrounded values in your calculator).
2. Step 2: Time of flight: it lands when y = 0 again, $0 = 15.0t - 4.9t^2$ , so $t = \frac{2v_{0y}}{g} = \frac{2(15.05)}{9.8} \approx 3.07$ s.
3. Step 3: Range: $x = (20.0)(3.07) \approx 61$ m.
4. Step 4: Maximum height: at the top $v_y = 0$ , so $h = \frac{v_{0y}^2}{2g} = \frac{15.0^2}{19.6} \approx 11.5$ m.
Answer: About 3.1 s, 61 m and 11.5 m.
Example 3
Differentiating a position vector
A particle's position is $\vec{r}(t) = (4t)\hat{i} + (6t - 5t^2)\hat{j}$ , in meters, with t in seconds. Find its velocity and acceleration vectors, the time when its velocity is purely horizontal, and its speed at t = 1.0 s.
Show the solutionHide the solution
1. Step 1: $\vec{v}(t) = 4\hat{i} + (6 - 10t)\hat{j}$ m/s and $\vec{a} = -10\hat{j}$ m/s², which is constant and downward: this is a projectile with g = 10 m/s².
2. Step 2: The velocity is horizontal when its y-component is zero: 6 − 10t = 0, so t = 0.60 s. That's the top of the path.
3. Step 3: At t = 1.0 s, $\vec{v} = 4\hat{i} - 4\hat{j}$ m/s, so the speed is $\sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.7$ m/s.
Answer: $\vec{v} = 4\hat{i} + (6 - 10t)\hat{j}$ m/s, $\vec{a} = -10\hat{j}$ m/s²; horizontal at t = 0.60 s; speed about 5.7 m/s at t = 1.0 s.

Common mistakes

Giving a projectile horizontal acceleration or a changing horizontal velocity. With no air resistance, $v_x$ stays constant.
Saying the velocity is zero at the top of the path. Only the vertical component is zero; the projectile still moves sideways.
Using the full launch speed in the vertical equations. Use the component $v_0\sin\theta$ , and $v_0\cos\theta$ horizontally.
Using level-ground shortcuts, like "45° gives the farthest range", when the launch and landing heights differ.

On the exam

Projectile questions often ask you to compare two launches, such as different angles or heights, and justify which lands first or travels farther. Reason with the separate x and y equations rather than intuition.
Sketching velocity components against time is a common task: $v_x$ is a flat line and $v_y$ is a straight line with slope −g.

Connected topics

Videos

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Check yourself

4 questions on 1.5 Motion in Two or Three Dimensions. Pick an answer to see if you got it, and why.

A ball is thrown horizontally at 15 m/s from the top of a cliff 20 m above level ground. Ignore air resistance and use g = 10 m/s².

Described situation

Question 1 of 4Calculator allowed

How long is the ball in the air?

Question 2 of 4Calculator allowed

How far from the base of the cliff does the ball land?

Question 3 of 4Calculator allowed

How fast is the ball moving just before it hits the ground?

Question 4 of 4Calculator allowed

On level ground with no air resistance, a ball launched at 25° above the horizontal lands 30 m away. At what other launch angle, with the same launch speed, would it land 30 m away?

0 of 4 answered