$\vec{V}$
A 2D vector. Has both magnitude and direction. The arrow above the symbol marks it as a vector.
Mistake Master
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Vectors and motion in two dimensions
2D motion looks intimidating, but it's really just Topic 1.2 done twice on perpendicular axes at once. Split the motion into $x$ and $y$ components, then handle each one with the 1D tools you already know.
§1
What 2D motion actually is.
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Any motion that doesn't run along a single straight line lives in two dimensions. A ball arcing through the air, a hiker walking northwest, a car turning a corner: each has both an east-west piece and a north-south piece. One signed number isn't enough to describe it.
The trick that makes 2D motion manageable is decomposition. Pick a coordinate system, usually $x$ horizontal and $y$ vertical. Resolve every vector (position, velocity, acceleration) into an $x$-component and a $y$-component. After that, the $x$-story and the $y$-story run on separate tracks. They share the clock, not the variables.
Projectile motion is the standard example. A ball thrown into the air has zero horizontal acceleration (nothing pushes it sideways once it leaves your hand) and a constant vertical acceleration of $g$ downward. So the $x$-story is constant velocity, and the $y$-story is constant acceleration. Two 1D problems, sharing only $t$.
§2
A 4-step procedure.
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The procedure below works for every 2D motion problem in the course. The four steps don't change; only the numbers do.
- Pick a coordinate system and name your axes. Usually $x$ horizontal, $y$ vertical, up positive. State which way is positive before you sign anything.
- Resolve every vector quantity into an $x$-component and a $y$-component. Position, velocity, acceleration each split into a pair. Use $V_x = V\cos\theta$ and $V_y = V\sin\theta$ when $\theta$ is measured from $+x$, or swap the trig if you measure $\theta$ from another axis.
- Solve the $x$-story and the $y$-story as separate 1D problems. Use the same kinematic equations from Topic 1.2, once for $x$ and once for $y$. The two stories share only the time variable $t$.
- Recombine if the question asks for a magnitude or angle. Magnitude: $|\vec{V}| = \sqrt{V_x^2 + V_y^2}$. Direction: $\theta = \tan^{-1}(V_y / V_x)$, with the quadrant checked from the signs of $V_x$ and $V_y$. Otherwise, the component answer is already the answer.
Step 3 is where the work happens, but step 1 is where most students lose points. A coordinate system you didn't write down is a coordinate system you'll forget halfway through.
§3
The pieces you'll meet.
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Symbols, relationships, and values that show up in §4, §5, and on the AP exam. Use this as a quick lookup.
$V_x$, $V_y$
The perpendicular components of $\vec{V}$ along the $x$ and $y$ axes. Each is a signed number; the sign carries the direction along that axis.
$|\vec{V}| = \sqrt{V_x^2 + V_y^2}$
Pythagorean theorem. Combines perpendicular components into a magnitude. Squaring removes the signs, so magnitude is never negative.
$\theta$
The angle between $\vec{V}$ and a reference axis. State which axis you're measuring from; the trig you use depends on it.
$V_x = V\cos\theta$
Projects onto the axis $\theta$ is measured from. If $\theta$ is from $+x$, this gives the $x$-component; if from $+y$, this gives the $y$-component instead.
$V_y = V\sin\theta$
Projects onto the axis perpendicular to the reference axis. Sine and cosine swap roles when you swap the reference axis.
$a_x = 0$
For an ideal projectile (no air resistance, no horizontal forces). Horizontal velocity stays at its launch value for the entire flight.
$a_y = -g$
For an ideal projectile, with up positive. Gravity is the only force in the air. Use $g = 10\ \text{m/s}^2$ for AP Physics 1.
§4
Worked example: horizontal launch from a tabletop.
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A ball rolls off the edge of a $1.25\ \text{m}$ tall table with horizontal velocity $v_x = 4\ \text{m/s}$. Find the time the ball spends in the air and where it lands relative to the edge of the table.
Step 1: coordinate system
Let $x$ be horizontal (positive in the direction of motion) and $y$ be vertical (positive up). Origin at the edge of the table.
Step 2: resolve into components
Initial velocity: $v_{0x} = +4\ \text{m/s}$, $v_{0y} = 0$ (the ball was rolling, not falling, just before launch).
Acceleration: $a_x = 0$ (nothing pushes horizontally once the ball leaves the table), $a_y = -10\ \text{m/s}^2$ (gravity, with up positive).
Initial position: $y_0 = +1.25\ \text{m}$. The landing position is $y = 0$.
Step 3: solve $x$ and $y$ separately
The $y$-story has constant acceleration and zero initial $y$-velocity: $$y = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2$$ Set $y = 0$ to find when the ball hits the ground: $$0 = 1.25 + 0 - 5t^2 \implies t^2 = 0.25 \implies t = 0.5\ \text{s}$$ Notice that $v_x$ never appeared in this calculation. The fall time depends only on $y$-quantities.
The $x$-story has constant velocity: $$x = v_{0x} t = (4)(0.5) = 2\ \text{m}$$
Step 4: recombine
The question asked for time aloft and range, both of which fell out of step 3. No recombination needed. The ball lands $2\ \text{m}$ from the edge of the table after $0.5\ \text{s}$ in the air.
§5
Worked example: launch at an angle.
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A ball is kicked from ground level with initial speed $v_0 = 20\ \text{m/s}$ at an angle of $53°$ above the horizontal. Find the maximum height the ball reaches and the total time it spends in the air.
Step 1: coordinate system
$x$ horizontal in the direction of the kick, $y$ vertical with up positive. Origin at the launch point.
Step 2: resolve into components
The angle is measured from $+x$, so $V_x = V\cos\theta$ and $V_y = V\sin\theta$: $$v_{0x} = (20)(0.6) = 12\ \text{m/s}, \quad v_{0y} = (20)(0.8) = 16\ \text{m/s}$$ Acceleration: $a_x = 0$, $a_y = -10\ \text{m/s}^2$.
Step 3: solve $x$ and $y$ separately
For maximum height, work in $y$ only. At the peak, $v_y = 0$: $$v_y = v_{0y} + a_y t \implies 0 = 16 - 10\, t_{\text{peak}} \implies t_{\text{peak}} = 1.6\ \text{s}$$ The height at that instant: $$\begin{aligned} y_{\text{peak}} &= v_{0y}\, t_{\text{peak}} + \dfrac{1}{2} a_y t_{\text{peak}}^2 \\ &= (16)(1.6) - 5(1.6)^2 = 25.6 - 12.8 = 12.8\ \text{m} \end{aligned}$$ For the total time of flight, the ball lands when $y = 0$ again. Since launch and landing heights are equal, the up trip and the down trip take the same time, so $t_{\text{total}} = 2\, t_{\text{peak}}$: $$t_{\text{total}} = (2)(1.6) = 3.2\ \text{s}$$
Step 4: recombine
Both quantities asked for came directly out of the $y$-story. No recombination needed for height or time. The horizontal range, if asked, would be $x = v_{0x} t_{\text{total}} = (12)(3.2) = 38.4\ \text{m}$.
Compare with §4: the procedure is identical. Step 1 is the same. Step 2 now actually uses trigonometry because the angle is nonzero. Step 3 is the same kinematic equations. The structure of 2D motion does not change with the launch angle; only the numbers in step 2 do.
§6
3 mistakes that cost real points.
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Each of these mistakes shows up reliably on the AP exam, in homework, and in graded labs. Each has a clear pattern: spot the shape of the wrong thinking, and you stop falling for it.
Pitfall · 01
"Going forward will slow the ball's fall."
This is the central error of 2D motion. It says: a ball that's also moving horizontally will fall slower than a ball just dropping, because the sideways motion somehow takes energy away from the fall. It's wrong. Once a projectile leaves the launcher, gravity pulls only along $y$. The $v_y$ equation never sees $v_x$, and the $v_x$ equation never sees $v_y$. A ball launched horizontally at $50\ \text{m/s}$ from a tabletop and a ball dropped straight down from the same height hit the ground at exactly the same instant.
Fix. When you solve any 2D problem, write $x$ on one line and $y$ on a separate line. Solve each like a 1D problem from Topic 1.2. They share only the time variable $t$; if a $v_x$ shows up in your $y$-equation, that's a flag.
Pitfall · 02
"At the peak of the arc, the ball is momentarily at rest."
At the top of a projectile's trajectory, the vertical velocity is zero. That much is true. The mistake is jumping from "$v_y = 0$" to "the ball is at rest" or "the acceleration is zero." Only $v_y$ is zero at the peak. $v_x$ is unchanged from launch (no force has touched it). Acceleration is still $g$ downward. That nonzero acceleration is exactly what flips $v_y$ from positive (rising) to negative (falling); if $a_y$ were zero, $v_y$ would just stay put.
Fix. Treat "$v = 0$" and "$a = 0$" as two independent statements. The first means the position is momentarily not changing; the second means the velocity is momentarily not changing. In a 1D vertical turnaround, the position is momentarily not changing. In 2D projectile motion, only the vertical component stops changing at the peak — $v_x$ remains nonzero, so the ball's position is still moving horizontally. In either case, $a = 0$ is the most false it gets: gravity is at its full strength.
Pitfall · 03
"$V_x$ is always $V\cos\theta$."
The mnemonic "$x$ goes with cosine, $y$ goes with sine" is correct, but only when $\theta$ is measured from the $+x$ axis. Cosine projects onto whichever axis you measured the angle from. If a problem says "the path makes an angle of $60°$ with north," then $\theta$ is measured from $+y$, and $V\cos\theta$ gives the $y$-component, not the $x$-component. Sine and cosine swap roles when you swap the reference axis. A student who memorizes "$x = V\cos\theta$" without checking the reference axis will use the wrong trig half the time.
Fix. Before grabbing trig, name the reference axis out loud or on the page: "the angle is measured from $+y$." Then cosine projects along that axis; sine projects perpendicular to it. The mnemonic is still useful, but only after the reference axis is locked in.
§7
Skill Check.
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Ten scenarios, in order. Pick from the chips and press Check my answer. The page remembers your work between visits.
0 OF 10 SCENARIOS COMPLETE
Resolve a vector along $+x$.
A velocity vector $\vec{V}$ has magnitude $20\ \text{m/s}$ and points at an angle of $37°$ above the $+x$ axis. What is its $x$-component? Use $\sin 37° = 0.6$ and $\cos 37° = 0.8$.
Magnitude
Direction
Reference axis: angle measured from north.
A drone's velocity vector $\vec{v}$ has magnitude $50\ \text{m/s}$ and points in a direction $30°$ east of north. (East is $+x$, north is $+y$.) Use $\sin 30° = 0.5$ and $\cos 30° = 0.866$. What is the $x$-component $v_x$?
Magnitude
Direction
Two perpendicular legs.
A robot drives $3\ \text{m}$ east, then $4\ \text{m}$ north, on level ground. What is the magnitude of its total displacement from the starting point?
Cliff drop: full velocity profile.
A ball is launched horizontally from the edge of an $80\ \text{m}$ tall cliff with initial speed $v_{0x} = 30\ \text{m/s}$. Use $g = 10\ \text{m/s}^2$. Answer each of the four parts below.
Part 1: vertical velocity at impact
Magnitude
Direction
Part 2: horizontal velocity at impact
Magnitude
Direction
Part 3: total time of flight
Time
Part 4: magnitude of resultant velocity at impact
Magnitude
Drop vs horizontal launch: which lands first?
From the edge of a $20\ \text{m}$ tall cliff, ball A is dropped from rest and ball B is launched horizontally with $v_0 = 15\ \text{m/s}$. Air resistance is negligible. Which ball reaches the ground first?
Velocity and acceleration at the peak.
A ball is launched from the ground with initial speed $v_0 = 20\ \text{m/s}$ at $53°$ above the horizontal, giving $v_{0x} = 12\ \text{m/s}$ and $v_{0y} = 16\ \text{m/s}$. Air resistance is negligible. At the highest point of the arc, what is the ball's velocity, and what is its acceleration?
Velocity at the peak
Magnitude
Direction
Acceleration at the peak
Magnitude
Direction
Just past the peak.
A ball is launched at an angle. It reaches its highest point at $t = 2.0\ \text{s}$. Air resistance is negligible. Use $g = 10\ \text{m/s}^2$. What is the ball's vertical velocity $v_y$ at $t = 2.2\ \text{s}$ (i.e., $0.2\ \text{s}$ after the peak)?
Magnitude
Direction
Symbolic: time aloft for a horizontal launch.
A ball is launched horizontally with speed $v_0$ from the edge of a cliff of height $H$. Air resistance is negligible. Express the time the ball spends in the air, in terms of $H$ and $g$.
Factor of change: quadruple the cliff height.
A ball is launched horizontally from a cliff with speed $v_0$ and travels a horizontal range $R$ before hitting the ground. If the cliff height is now quadrupled (so the new height is $4H$), but $v_0$ stays the same, what is the new range?
Symbolic: peak height of an angled launch.
A projectile is launched from ground level with initial speed $v_0$ at an angle $\theta$ above the horizontal. Air resistance is negligible. Express the maximum height the projectile reaches, in terms of $v_0$, $\theta$, and $g$.
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