AP Physics C: Mechanics · Topic 1.5

Topic 1.5: Motion in Two or Three Dimensions

Unit 1: Kinematics10-15% of the multiple-choice section

Topic 1.5 says motion in two or three dimensions can be analysed with one-dimensional relationships once it is separated into components, and that each component may have its own nonuniform acceleration. Its boundary statement limits Mechanics to calculating in two dimensions.

AP Physics: Unit 1 (topics 1.5 Motion in Two or Three Dimensions). AP Physics C: Mechanics Unit 1, Topic 1.5. One learning objective, 1.5.A: describe the motion of an object moving in two or three dimensions. Four essential-knowledge statements, none with sub-statements. 1.5.A.1 motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. 1.5.A.2 velocity and acceleration may be different in each dimension and may be nonuniform. 1.5.A.3 motion in one dimension may be changed without causing a change in a perpendicular dimension. 1.5.A.4 projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension. NO equations are printed for this topic. ONE boundary statement, quoted whole: 'AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions.' Suggested skills 1.B, 2.A, 2.D, 3.A, 3.C, which is the ONLY Unit 1 topic whose suggested-skill list is identical to its AP Physics 1 counterpart's. This is one of the four C: Mechanics topics whose title differs from its AP Physics 1 sibling: AP Physics 1's Topic 1.5 is 'Vectors and Motion in Two Dimensions' and has TWO learning objectives. Its 1.5.A covers resolving a vector into perpendicular components (EK 1.5.A.1 to 1.5.A.3, printing sin theta = a/c, cos theta = b/c, tan theta = a/b and a^2 + b^2 = c^2 beside a right-triangle figure), work that AP Physics C relocated to Topic 1.1 and rewrote with unit vectors; its 1.5.B covers the motion, with 1.5.B.1 matching Physics C's 1.5.A.1 except for 'two' in place of 'two or three', and 1.5.B.2 matching Physics C's 1.5.A.4 word for word. AP Physics C's 1.5.A.2 and 1.5.A.3 have no AP Physics 1 counterpart at all, and AP Physics 1's Topic 1.5 prints no boundary statement. Consistent with the boundary statement, 1.1.A.4.i introduces i-hat, j-hat and k-hat while the component-addition line at 1.1.A.4.iii and on the Table of Information's Vectors table stops at j-hat. The exam-conventions box on that appendix page states that air resistance is assumed to be negligible unless otherwise stated, which is what keeps 1.5.A.4's 'constant' acceleration constant.

What Topic 1.5 requires

One learning objective, 1.5.A: describe the motion of an object moving in two or three dimensions. Four essential-knowledge statements, none with sub-statements.

1.5.A.1 Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
1.5.A.2 Velocity and acceleration may be different in each dimension and may be nonuniform.
1.5.A.3 Motion in one dimension may be changed without causing a change in a perpendicular dimension.
1.5.A.4 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.

Suggested skills: 1.B create quantitative graphs with appropriate scales and units, including plotting data; 2.A derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D predict new values or factors of change of physical quantities using functional dependence between variables; 3.A create experimental procedures that are appropriate for a given scientific question; and 3.C justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. This is the one topic in Unit 1 whose suggested skills are identical to its AP Physics 1 counterpart's, all five of them.

Topic 1.5 prints no equations. Everything it needs was printed earlier: the unit-vector notation at 1.1.A.4.i, the derivative and integral forms at 1.2.C.1 and 1.3.A.4, and the three constant-acceleration equations at 1.3.A.2 with their note that they apply in any single dimension.

One boundary statement, quoted in full below.

Why the title differs from the AP Physics 1 topic

This is one of only four C: Mechanics topics whose title is not shared with an AP Physics 1 topic. AP Physics 1's Topic 1.5 is called Vectors and Motion in Two Dimensions; this one is Motion in Two or Three Dimensions. The word "Vectors" is gone from the front and "or Three" has been added at the back, and both edits are structural.

AP Physics 1's Topic 1.5 has two learning objectives. Its 1.5.A is entirely about resolving a vector into perpendicular components, and it prints four equations to do it with: $\sin\theta = a/c$ , $\cos\theta = b/c$ , $\tan\theta = a/b$ and $a^2 + b^2 = c^2$ , next to a labelled right-triangle figure. Only its second objective, 1.5.B, is about the motion.

AP Physics C moved the component resolution to Topic 1.1, where it is done with unit vectors rather than with a triangle. So this topic has one objective instead of two, and it is purely about the motion. That is also why it prints no equations while the algebra-based version prints four: this course did the vector algebra three topics earlier.

Mapping the statements across:

AP Physics C	AP Physics 1	Relationship
in Topic 1.1, at 1.1.A.4	1.5.A.1, 1.5.A.2, 1.5.A.3	resolving components, relocated and rewritten with $\hat{i}$ and $\hat{j}$
1.5.A.1	1.5.B.1	same sentence, with "two or three" replacing "two"
1.5.A.2	nothing	new
1.5.A.3	nothing	new
1.5.A.4	1.5.B.2	identical text
boundary statement	none	new

The boundary statement, quoted whole

The title says two or three dimensions. The boundary statement says which of those you will be asked to calculate in:

"AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions."

Both sentences matter and they say different things about different exams.

For Mechanics: quantitative work is two-dimensional. Three-dimensional notation appears, and you should be able to read it, but the calculations you are set will resolve into two axes. This is consistent across the whole framework. Statement 1.1.A.4.i introduces all three basis vectors $\hat{i}$ , $\hat{j}$ and $\hat{k}$ , and then the component-addition line printed at 1.1.A.4.iii and on the Table of Information's Vectors table stops at $\hat{j}$ .

For Electricity and Magnetism: the same statement extends the expectation to a qualitative description of three-dimensional particle motion. That is not idle. A charge moving in a magnetic field experiences a force perpendicular to both its velocity and the field, which is a genuinely three-dimensional situation, and describing the resulting helical path in words is exactly the qualitative task this sentence licenses.

AP Physics 1's Topic 1.5 prints no boundary statement at all. Its restriction to two dimensions is built into its title and its statements instead.

The two statements AP Physics 1 does not have

Statements 1.5.A.2 and 1.5.A.3 have no counterpart anywhere in AP Physics 1's Topic 1.5. Between them they say what the C course is allowed to do with two-dimensional motion.

1.5.A.2: velocity and acceleration may be different in each dimension and may be nonuniform.

The last three words are the ones that matter, and they are the Topic 1.3 boundary lift reappearing in two dimensions. Once each component may be nonuniform, a two-dimensional motion is no longer necessarily a projectile. It might have a constant acceleration along $x$ and a time-varying one along $y$ , and then the $x$ motion obeys the three kinematic equations and the $y$ motion does not. A worked example below is exactly that case. AP Physics 1's Topic 1.5 contains nothing that permits this, and its Topic 1.3 boundary statement forbids the quantitative version of it.

The practical rule that follows: decide constant or not for each axis separately. There is no such thing as "the acceleration is constant" in a two-dimensional problem until you have asked the question once per component.

1.5.A.3: motion in one dimension may be changed without causing a change in a perpendicular dimension.

This is component independence stated as physics rather than as algebra. It is the reason 1.5.A.1 works at all: if changing $x$ could disturb $y$ , you could not split the problem into two one-dimensional problems and solve them separately.

It has a clean mathematical statement in this course. Write $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$ and differentiate; because $\hat{i}$ and $\hat{j}$ are fixed perpendicular directions, $x(t)$ and $y(t)$ evolve under separate equations that share only the variable $t$ . The whole coupling between the two motions is the clock.

That is also the practical solving strategy, and it is worth writing out because it is what 1.5.A.1 is telling you to do:

Choose axes and state them.
Resolve every given quantity onto them, using the unit-vector machinery from Topic 1.1.
Solve each axis as a separate one-dimensional problem, checking per axis whether the acceleration is constant.
Recombine only at the end, and only when the question asks for a magnitude or a direction.

Step 3 is where the C course diverges: on an axis with a nonuniform acceleration, the one-dimensional relationship you apply is a derivative or an integral rather than one of the three printed equations.

Doing the calculus one component at a time

Statement 1.5.A.1's phrase "one-dimensional kinematic relationships" covers every relationship Unit 1 has printed, not only the three constant-acceleration equations. In this course that means the derivative and integral forms come apart per component too:

\vec{v} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} \qquad \vec{a} = \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j}

and going the other way, with a separate constant of integration for every component at every step. A two-dimensional problem that starts from an acceleration therefore needs four initial conditions, not two: $v_{x0}$ , $v_{y0}$ , $x_0$ and $y_0$ .

Two derived quantities are worth having, because questions ask for them and neither is a component:

Speed is the magnitude of the velocity vector, $\left| \vec{v} \right| = \sqrt{v_x^2 + v_y^2}$ . It is smallest when the components are individually smallest, which is not generally when either one is zero.
The trajectory is the path in space, $y$ as a function of $x$ with time eliminated. Getting it is a skill 2.A exercise: solve one component for $t$ and substitute into the other. For a projectile this produces the parabola, and a worked example below derives one.

A detail that separates a correct answer from a nearly correct one: an object's velocity is always tangent to its trajectory, but its acceleration is not generally along the trajectory at all. If you have found the direction of $\vec{v}$ , you have found the direction of travel; you have not found the direction of $\vec{a}$ .

Projectile motion, the special case at 1.5.A.4

Statement 1.5.A.4 defines it precisely, and the definition is worth reading as a definition rather than as a scenario: projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.

So the word "projectile" in this framework is not about gravity or about throwing things. It names a pattern of accelerations. Any two-dimensional motion matching that pattern is a projectile problem, including a charged particle crossing a uniform electric field, which is why the definition earns its place in a course whose second half is electromagnetism.

Read against 1.5.A.2, it is the special case where both components happen to be uniform, one of them trivially so. That is why every projectile question can be done with the three printed equations, and why most two-dimensional questions in this course that are not projectiles cannot be.

	The zero-acceleration axis	The constant-acceleration axis
Acceleration	$0$	constant and nonzero
Velocity component	constant	changes linearly with $t$
Position component	linear in $t$	quadratic in $t$
Available relationships	$x = x_0 + v_{x0}t$	all three from 1.3.A.2

The two axes share the clock and nothing else, which is 1.5.A.3 again. The time to fall does not depend on the horizontal speed, and the horizontal distance covered does not depend on the vertical drop except through the shared $t$ . Time is the variable to solve for first in almost every projectile question.

One caution about scope. Statement 1.5.A.4 says constant acceleration in the second dimension, so it describes the no-air-resistance case, and the Table of Information's exam-conventions box backs that up: air resistance is assumed to be negligible unless otherwise stated. When a question does state otherwise, the vertical acceleration stops being constant, and you are in Topic 2.9, Resistive Forces, a topic AP Physics 1 does not have. For the standard routine, the guide on projectile motion problems covers the procedure.

Traps on this topic

Assuming both components have constant acceleration. Statement 1.5.A.2 explicitly permits nonuniform, and per component. Check each axis before reaching for the three equations.

Mixing components in one equation. A kinematic equation applies along one axis with that axis's own quantities. Putting $v_{y0}$ and $a_x$ into the same equation is not a slip, it is a category error.

Thinking the launch speed appears in the vertical equations. It appears as $v_{0}\sin\theta$ . The speed itself belongs to neither axis.

Looking for zero speed at the top of a trajectory. At the highest point $v_y = 0$ , but $v_x$ is unchanged, so the speed there is $v_x$ and generally nonzero. It is the minimum speed of the flight, not zero.

Assuming minimum speed happens when a component is zero. True for a projectile because one component is constant. Not true in general: with both components varying, minimising $v_x^2 + v_y^2$ is its own calculation.

Reporting the direction of $\vec{a}$ as the direction of motion. The velocity is tangent to the path; the acceleration usually is not.

Carrying only two initial conditions into a two-dimensional integration. Four are needed, two per component.

Reading the boundary statement as a ban on three dimensions. It says AP Physics C: Mechanics only expects quantitative analysis in two, and that AP Physics C: Electricity and Magnetism expects a qualitative description in three. Three-dimensional notation appears from 1.1.A.4.i onwards.

If you want the algebra-based version of this topic

The titles differ here, which is the clue that the topics differ. AP Physics 1's Topic 1.5 does two jobs, resolving vectors and analysing two-dimensional motion. AP Physics C's does one, because the first job was done in Topic 1.1.

	AP Physics 1 Topic 1.5	AP Physics C Topic 1.5
Title	Vectors and Motion in Two Dimensions	Motion in Two or Three Dimensions
Learning objectives	1.5.A resolving components, 1.5.B the motion	1.5.A the motion only
Resolving components	here, with a right-triangle figure	in Topic 1.1, with unit vectors
Equations printed	four, the three trig ratios and Pythagoras	none
Nonuniform per-component acceleration	not stated	1.5.A.2
Component independence	not stated	1.5.A.3
Projectile definition	1.5.B.2	1.5.A.4, identical text
Boundary statement	none	quantitative work limited to two dimensions
Suggested skills	1.B, 2.A, 2.D, 3.A, 3.C	the same five

The identical skill list is worth noticing. Both courses want the same things done here, plotting quantitative graphs, deriving symbolic expressions, predicting by functional dependence, designing procedures and justifying claims. What differs is the class of motion those skills are pointed at.

If you are studying for AP Physics 1, the page you want is Vectors and Motion in Two Dimensions, and the projectile motion calculator serves both courses. If you are studying for AP Physics C: Mechanics, this is the last topic of Unit 1; the unit hub ties the five together, and Unit 2 opens with systems and centre of mass.

One component uniform, the other not

A particle moves in the $xy$ plane with $\vec{r}(t) = \left( 3.0t^2 \right)\hat{i} + \left( 8.0t - 0.50t^3 \right)\hat{j}$ , in metres with $t$ in seconds. Find $\vec{v}(t)$ and $\vec{a}(t)$ , the velocity and acceleration at $t = 2.0$ s, and the instant and position at which the particle is farthest from the $x$ -axis. State which component obeys the three kinematic equations.

Differentiate each component. $\vec{v} = 6.0t\,\hat{i} + \left( 8.0 - 1.5t^2 \right)\hat{j}$ m/s.
Differentiate again. $\vec{a} = 6.0\,\hat{i} - 3.0t\,\hat{j}$ m/s $^2$ .
Read off 1.5.A.2 in the result. $a_x = 6.0 \ \mathrm{m/s^2}$ is constant, so the $x$ motion obeys the three equations printed at 1.3.A.2. $a_y = -3.0t$ is not constant, so the $y$ motion does not, and the $y$ component had to be handled by differentiation.
At $t = 2.0$ s. $\vec{v} = 12\hat{i} + \left( 8.0 - 6.0 \right)\hat{j} = 12\hat{i} + 2.0\hat{j}$ m/s, so the speed is $\sqrt{144 + 4.0} = 12.2$ m/s at $\tan^{-1}(2.0/12) = 9.5^\circ$ above the $+x$ axis.
Acceleration at the same instant. $\vec{a} = 6.0\hat{i} - 6.0\hat{j}$ m/s $^2$ , magnitude $\sqrt{72} = 8.49$ m/s $^2$ , pointing $45^\circ$ below the $+x$ axis. Note it is nowhere near parallel to $\vec{v}$ .
Farthest from the $x$ -axis is where $v_y = 0$ . $8.0 - 1.5t^2 = 0$ gives $t^2 = 5.333$ and $t = 2.31$ s.
Position there. $y = 8.0(2.31) - 0.50(2.31)^3 = 18.48 - 6.16 = 12.3$ m, and $x = 3.0(2.31)^2 = 16.0$ m.

$\vec{v} = 6.0t\hat{i} + (8.0 - 1.5t^2)\hat{j}$ m/s and $\vec{a} = 6.0\hat{i} - 3.0t\hat{j}$ m/s $^2$ . At $t = 2.0$ s the speed is 12.2 m/s at $9.5^\circ$ and the acceleration is 8.49 m/s $^2$ at $45^\circ$ below the $+x$ axis. The particle reaches $y = 12.3$ m at $t = 2.31$ s, at $x = 16.0$ m. Only the $x$ component has constant acceleration.

Independence of perpendicular components, and the trajectory

A puck slides across a frictionless horizontal table in the $+x$ direction at a steady 2.4 m/s. At $t = 0$ , from the position $x = 0$ , $y = 0$ , a constant sideways push begins, giving it $a_y = 3.0 \ \mathrm{m/s^2}$ and leaving $a_x = 0$ . Find the value of $y$ when $x = 6.0$ m, the $y$ velocity component then, and a symbolic expression for the trajectory $y(x)$ .

Classify it against 1.5.A.4. Zero acceleration along $x$ , constant nonzero acceleration along $y$ : this is projectile motion by the CED's definition, even though the table is horizontal and gravity is not involved.
Use 1.5.A.3. The sideways push is entirely along $y$ , so it cannot change the $x$ motion. The $x$ component stays at 2.4 m/s throughout.
Time to reach $x = 6.0$ m. $t = 6.0/2.4 = 2.5$ s.
The $y$ displacement in that time, from $y = y_0 + v_{y0}t + \frac{1}{2}a_yt^2$ with $y_0 = 0$ and $v_{y0} = 0$ . $y = \frac{1}{2}(3.0)(2.5)^2 = \frac{1}{2}(3.0)(6.25) = 9.4$ m.
The $y$ velocity there. $v_y = a_yt = (3.0)(2.5) = 7.5$ m/s, so the speed is $\sqrt{(2.4)^2 + (7.5)^2} = 7.9$ m/s.
The trajectory, eliminating $t$ per skill 2.A. From the $x$ equation, $t = x/v_{x0}$ . Substituting, $y = \frac{1}{2}a_y\left( \dfrac{x}{v_{x0}} \right)^2 = \dfrac{a_y}{2v_{x0}^2}x^2$ .
Check numerically. $\dfrac{3.0}{2(2.4)^2} = \dfrac{3.0}{11.52} = 0.2604$ m $^{-1}$ , so $y = 0.2604x^2$ , and at $x = 6.0$ m that gives $y = 0.2604(36) = 9.4$ m, matching step 4.

$y = 9.4$ m when $x = 6.0$ m, with $v_y = 7.5$ m/s and a speed of 7.9 m/s. The trajectory is the parabola $y = \dfrac{a_y}{2v_{x0}^2}x^2 = 0.2604x^2$ , with $x$ and $y$ in metres.

Integrating an acceleration vector, component by component

A particle starts from rest at the origin with acceleration $\vec{a}(t) = 2.0\hat{i} + \left( 6.0 - 4.0t \right)\hat{j}$ , in m/s $^2$ with $t$ in seconds. Find $\vec{v}(t)$ and $\vec{r}(t)$ , and the position and speed at the instant the particle stops moving away from the $x$ -axis.

Integrate each component once. $v_x = \int 2.0\,dt = 2.0t + C$ , and $v_x(0) = 0$ gives $C = 0$ . Likewise $v_y = 6.0t - 2.0t^2$ , using $v_y(0) = 0$ . So $\vec{v} = 2.0t\,\hat{i} + \left( 6.0t - 2.0t^2 \right)\hat{j}$ m/s.
Integrate again, with $x(0) = y(0) = 0$ . $\vec{r} = 1.0t^2\,\hat{i} + \left( 3.0t^2 - \frac{2}{3}t^3 \right)\hat{j}$ m. That is four constants of integration fixed by four initial conditions, two per component.
The particle stops moving away from the $x$ -axis when $v_y = 0$ . $6.0t - 2.0t^2 = 2.0t(3.0 - t) = 0$ , so $t = 0$ at the start and $t = 3.0$ s after it.
Position at $t = 3.0$ s. $x = 1.0(9.0) = 9.0$ m and $y = 3.0(9.0) - \frac{2}{3}(27) = 27.0 - 18.0 = 9.0$ m.
Velocity there. $v_x = 2.0(3.0) = 6.0$ m/s and $v_y = 0$ , so the speed is 6.0 m/s, directed along $+x$ .
Check against 1.5.A.2. Both components had different accelerations, and the $y$ one was nonuniform while the $x$ one was constant, so neither the motion as a whole nor either component could be assumed uniform without checking.

$\vec{v} = 2.0t\hat{i} + (6.0t - 2.0t^2)\hat{j}$ m/s and $\vec{r} = 1.0t^2\hat{i} + \left( 3.0t^2 - \frac{2}{3}t^3 \right)\hat{j}$ m. At $t = 3.0$ s the particle is at $(9.0, 9.0)$ m moving at 6.0 m/s parallel to the $x$ -axis.

Frequently asked questions

What is the difference between AP Physics C Topic 1.5 and AP Physics 1 Topic 1.5?

Even the titles differ: AP Physics 1's is Vectors and Motion in Two Dimensions and AP Physics C's is Motion in Two or Three Dimensions. AP Physics 1 spends its first learning objective on resolving vectors into perpendicular components with trigonometry and prints four equations for it; AP Physics C moved that work into Topic 1.1 and does it with unit vectors, so this topic has one objective and prints no equations. AP Physics C also adds two statements AP Physics 1 has no version of: that velocity and acceleration may be different in each dimension and may be nonuniform, and that motion in one dimension may be changed without causing a change in a perpendicular dimension.

Does AP Physics C Mechanics test motion in three dimensions?

Not quantitatively. The Topic 1.5 boundary statement reads that AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions, and that AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions. Three-dimensional notation still appears, since essential knowledge 1.1.A.4.i introduces the k-hat basis vector, but the component-addition equation printed on the equation sheet stops at j-hat.

Can acceleration be different in each direction in AP Physics C?

Yes, and it can also vary with time in one direction while staying constant in the other. Essential knowledge 1.5.A.2 states that velocity and acceleration may be different in each dimension and may be nonuniform. AP Physics 1's Topic 1.5 contains no equivalent statement, and its Topic 1.3 boundary statement blocks quantitative analysis of nonuniform acceleration entirely. The practical rule in AP Physics C is to decide constant or not for each axis separately: the three kinematic equations may be legal along one axis and illegal along the perpendicular one in the same problem.

Why can two-dimensional motion be split into two one-dimensional problems?

Because perpendicular components are independent. Essential knowledge 1.5.A.3 states that motion in one dimension may be changed without causing a change in a perpendicular dimension, and 1.5.A.1 states that motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. Mathematically, writing the position as x of t along i-hat plus y of t along j-hat and differentiating gives two separate equations, because the basis directions are fixed and perpendicular. The only thing the two motions share is the clock.

How does AP Physics C define projectile motion?

Essential knowledge 1.5.A.4 defines it as a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension. The definition is about the pattern of accelerations rather than about gravity, so any motion matching it counts, including a charged particle crossing a uniform electric field. AP Physics 1 gives exactly the same sentence, at its 1.5.B.2. Because both components are uniform, every projectile problem can be solved with the three constant-acceleration equations, one axis at a time.

How many initial conditions does a two-dimensional integration need?

Four, two for each component. Integrating an acceleration component once gives a velocity component up to a constant, fixed by that component's initial velocity, and integrating again gives a position component up to a second constant, fixed by that component's initial position. So a problem that hands you an acceleration vector and asks for a position vector needs the initial values of v-sub-x, v-sub-y, x and y. Losing one produces an answer with the right functional shape and the wrong values.

Is the speed at the top of a projectile's path zero?

No. At the highest point the vertical component of velocity is zero, but the horizontal component is unchanged, because essential knowledge 1.5.A.3 says a change in one dimension does not affect a perpendicular one. The speed at the top therefore equals the horizontal component, which is the smallest speed of the whole flight but is not zero unless the launch was straight up. The acceleration at that point is unchanged too, still the full constant value in the vertical direction.