4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The diagram (not to scale) shows a ball kicked horizontally off the top of a vertical cliff above level ground. Air resistance is negligible. Use g = 10 m/s². How far from the base of the cliff does the ball land?
Answer and reasoning
A24 m A student who uses the launch speed as the initial vertical velocity picks this: 45 = 12t + 5t² gives t ≈ 2.0 s and 12 × 2.0 ≈ 24 m. The launch velocity is horizontal, so its vertical component is zero.
B54 m A student who thinks the ball falls 10 m in each second picks this: 45 m ÷ 10 m/s = 4.5 s, and 12 × 4.5 = 54 m. The ball accelerates downward, falling 5 m in the first second, 15 m in the next and 25 m in the third: 45 m in 3.0 s.
C18 m A student who thinks the ball's forward motion is used up during the flight picks this: if vx fell steadily from 12 m/s to zero, the average would be 6 m/s, and 6 × 3.0 = 18 m. A projectile has ax = 0, so vx stays 12 m/s.
D36 mCorrect Treat the components separately. Vertically the ball starts with vy = 0 and falls 45 m: 45 = (1/2)(10)t², so t = 3.0 s. Horizontally nothing changes vx = 12 m/s, so Δx = (12 m/s)(3.0 s) = 36 m.
Working Vertical: 45 = (1/2)(10)t² → t = 3.0 s. Horizontal: vx = 12 m/s constant → Δx = 12 × 3.0 = 36 m.
The figure shows graphs of the x and y components of the velocity of an object moving in the xy plane, as functions of time t. Which statement about the object's acceleration from t = 0 to t = 3 s is correct?
Answer and reasoning
AIt is constant and directed in the −x directionCorrect Each acceleration component is the slope of the matching velocity graph. The vx graph is a straight line with slope (0 − 6)/3 = −2 m/s², and the vy graph is horizontal, so ay = 0. The acceleration is a constant 2 m/s² in the −x direction.
BIt points along the object's velocity at every instant A student who thinks the acceleration points along the direction of motion picks this. The velocity turns from (6, 4) m/s to (0, 4) m/s, while the acceleration stays in the −x direction, at an angle to the velocity.
CIt is greatest at t = 0, when vx is greatest A student who reads the acceleration from the height of the velocity graph picks this. The acceleration is the slope of the vx graph, which is the same at every instant: vx is largest at t = 0, but it changes at a constant rate.
DIt has a +y component, since vy is not zero A student who thinks a velocity component must come with an acceleration component picks this. vy is a constant 4 m/s, so it does not change: ay = 0.
Working ax = slope of vx–t = (0 − 6)/3 = −2 m/s², constant; ay = slope of vy–t = 0. So a⃗ = 2 m/s² in −x, constant, while v⃗ turns from (6, 4) to (0, 4) m/s.
A ball rolls off the edge of a horizontal table 0.80 m high at 1.5 m/s and lands a horizontal distance D from the edge of the table. The same ball then rolls off the edge of a different horizontal table, 1.8 m high, at 3.0 m/s. Air resistance is negligible. It now lands a horizontal distance kD from the edge. What is k?
Answer and reasoning
A4.5 A student who thinks a falling object covers equal distances in equal times, so that the fall time is proportional to the height, picks this: 2 × (1.8/0.80) = 4.5. The ball accelerates downward, so t grows only as √h, by a factor of 1.5.
B3.0Correct The fall time depends only on the height: t = √(2h/g), so it grows by √(1.8/0.80) = 1.5. The horizontal velocity is unchanged during each fall and is doubled from 1.5 to 3.0 m/s. Δx = vx t grows by 2 × 1.5 = 3.0.
C2.0 A student who thinks independent motions mean the landing distance depends only on the horizontal speed picks this. The horizontal and vertical motions are independent but last the same time, and the higher table gives a fall 1.5 times as long.
D4.0 A student who applies the level-ground range result, in which the range depends only on the launch speed and angle and grows as its square, picks this: (3.0/1.5)² = 4. For a horizontal launch Δx = vx t is proportional to vx, and the greater height also lengthens t.
Working t = √(2h/g) ∝ √h; Δx = vx t ∝ vx√h. k = (3.0/1.5) × √(1.8/0.80) = 2 × 1.5 = 3.0.
A ball is launched over level ground with speed v₀ at an angle θ above the horizontal. Air resistance is negligible. Which describes the ball's velocity at the highest point of its path?
Answer and reasoning
AIt is zero, since the ball stops for an instant at the top A student who thinks every projectile is momentarily at rest at the top picks this. Only vy is zero there; the ball is still moving horizontally at v₀ cos θ.
BIt is less than v₀ cos θ and is still decreasing A student who thinks the ball's forward motion is gradually used up picks this. With air resistance negligible, vx does not change, so at the top it still equals v₀ cos θ, and vy = 0 there.
CIt is horizontal and equal to v₀ cos θ, as at launchCorrect At the top only the vertical component of velocity is zero. A projectile's acceleration is vertical (ax = 0), so vx keeps its launch value v₀ cos θ. The velocity at the top is therefore horizontal with magnitude v₀ cos θ.
DIt is horizontal and equal to v₀, the launch speed A student who treats the whole launch speed as the horizontal velocity picks this, since at the top all of the motion is horizontal. Only the horizontal component of the launch velocity, v₀ cos θ, survives at the top; the vertical component v₀ sin θ has been reduced to zero.
Working At the top vy = 0 and vx = v₀ cos θ (ax = 0, so unchanged since launch): velocity horizontal, magnitude v₀ cos θ.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.5.A.1 Components of motion Fix
Components of motion
A two-dimensional motion can be described by its x and y components, x(t) and y(t), with vx = dx/dt, vy = dy/dt, ax = dvx/dt and ay = dvy/dt. Each component obeys one-dimensional kinematics, and the components share the same time t.
Resolving a launch velocity
A velocity of magnitude v₀ at angle θ above the horizontal has components vx0 = v₀ cos θ and vy0 = v₀ sin θ. Each component, not v₀ itself, is used in the equations for its own dimension.
Students often think The launch speed v₀ can be used as the initial velocity in the vertical (or horizontal) equations, without resolving it into components. In fact No. The launch velocity must first be resolved into components: vy0 = v₀ sin θ for the vertical equations and vx0 = v₀ cos θ for the horizontal ones (for a horizontal launch, vy0 = 0).
Students often think The horizontal component is v₀ sin θ and the vertical component is v₀ cos θ, or the two functions can be used either way round. In fact For an angle θ measured from the horizontal, the horizontal component is v₀ cos θ and the vertical component is v₀ sin θ; for an angle measured from the vertical, the functions swap.
1.5.A.2 Independent components Fix
Independent components
The velocity and acceleration can differ in each dimension: an object can have ax ≠ 0 and ay = 0, constant vy and changing vx, or a nonuniform acceleration in one dimension only.
Students often think An object's acceleration points along its velocity (its direction of motion) at every instant. In fact Not in general. The acceleration points along the change in velocity. In two dimensions the acceleration can be at any angle to the velocity; for a projectile it is always downward while the velocity turns.
Students often think The value of the acceleration can be read from the height of the velocity graph, so the acceleration is greatest when the velocity is greatest. In fact No. The acceleration is the slope of the velocity–time graph, not its height. A velocity component that is large but changing at a steady rate has a constant acceleration.
1.5.A.3 Independence of perpendicular motions Fix
Independence of perpendicular motions
A change in the motion in one dimension does not change the motion in a perpendicular dimension. For example, giving a projectile a larger horizontal velocity does not change its vertical motion or how long it takes to fall.
Students often think A falling object covers equal distances in equal times (with g ≈ 10 m/s², about 10 m in each second), so the time to fall is proportional to the height. In fact No. A falling object accelerates: its velocity changes by about 10 m/s each second, and the distance it covers in successive seconds grows (5 m, 15 m, 25 m, … from rest). The time to fall from rest grows as the square root of the height.
Students often think Because the horizontal and vertical motions are independent, the horizontal distance a projectile travels depends only on its horizontal velocity, not on the height from which it is launched. In fact Yes. The horizontal and vertical motions are independent, but they last the same time. A greater launch height gives a longer fall, and the constant horizontal velocity acts for that longer time.
1.5.A.4 Projectile Fix
Projectile
An object moving under gravity alone (air resistance negligible) near Earth's surface: ax = 0, so vx is constant, and ay = −g (up positive), constant and nonzero, so its path is a parabola (a vertical straight line if vx = 0).
Time of flight
The time a projectile spends in the air, determined by its vertical motion alone (initial vy, launch height and g), whatever its horizontal velocity. Unit: s.
Range (horizontal distance)
The horizontal displacement of a projectile between launch and landing: Δx = vx0 t, where t is the time of flight. Unit: m.
Highest point of a trajectory
The point where vy = 0. The velocity there is horizontal and equal to vx0, and the acceleration is still g downward.
Students often think A projectile carries 'forward motion' from its launch that is gradually used up, so its horizontal velocity decreases during the flight (and has run down by the time it lands). In fact No, not if air resistance is negligible. A projectile has zero horizontal acceleration, ax = 0, so vx stays equal to its launch value from launch to landing.
Students often think At the highest point of its path, any projectile is momentarily at rest. In fact Only if it was launched straight up. At the highest point vy = 0, but vx keeps its launch value, so the velocity there is horizontal and nonzero.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
A ball is launched from level ground with speed v₀ at 37° above the horizontal (sin 37° = 0.60, cos 37° = 0.80). Air resistance is negligible, and the acceleration due to gravity is g. What is the maximum height the ball reaches?
Answer and reasoning
A0.18 v₀²/gCorrect Only the vertical component matters for the height: vy0 = v₀ sin 37° = 0.60v₀. At the top vy = 0, so 0 = (0.60v₀)² − 2gH and H = 0.36v₀²/(2g) = 0.18 v₀²/g.
B0.32 v₀²/g A student who takes the vertical component as v₀ cos 37° = 0.80v₀ picks this: (0.80v₀)²/(2g) = 0.32 v₀²/g. With the angle measured from the horizontal, the vertical side is opposite the angle, so vy0 = v₀ sin 37°.
C0.50 v₀²/g A student who uses the launch speed v₀ as the initial vertical velocity picks this: v₀²/(2g). That is the height for a launch straight up; at 37° only 0.60v₀ of the launch velocity is vertical.
D0.36 v₀²/g A student who multiplies the initial vertical velocity by the time to the top picks this: (0.60v₀)(0.60v₀/g) = 0.36 v₀²/g. The vertical velocity falls steadily to zero, so the average is half of 0.60v₀, which gives 0.18 v₀²/g.
Working vy0 = v₀ sin 37° = 0.60v₀. At the top vy = 0: 0 = (0.60v₀)² − 2gH → H = 0.36v₀²/(2g) = 0.18 v₀²/g.
The diagram shows the paths of three balls, A, B and C, launched from the origin on level ground, and the horizontal component vx of each ball's velocity. Air resistance is negligible. Which ranks the times the balls spend in the air, from longest to shortest?
Answer and reasoning
AB > C > A A student who thinks the ball that lands farthest away must have been in the air longest picks this, ranking by landing distance. B goes farthest because of its larger horizontal velocity, not because it is in the air longer; A, with the highest arc, stays up longest.
BC > B > A A student who thinks a larger horizontal velocity holds a ball up picks this, ranking by vx. The horizontal motion has no effect on the vertical motion; the time in the air is set by how high each ball rises.
CA > B > CCorrect The time in the air depends only on the vertical motion. A ball that rises higher had a larger upward velocity component and takes longer to rise and to fall back: A (about 11 m) stays up longest, then B (5 m), then C (about 1 m). The horizontal components do not affect the times.
DA = B = C A student who thinks all projectiles launched from the same height land together picks this. That is true only for equal initial vertical velocities; these balls rise to different heights, so they have different vertical velocities and different times in the air.
Working Time of flight from vertical motion only: H = vy0²/(2g), T = 2vy0/g. A rises highest (≈11 m: T = 3.0 s), B 5 m (2.0 s), C ≈1 m (1.0 s). A > B > C. (Ranges: B 28 > C 20 > A 18 m; vx: C > B > A.)
A ball is thrown from the top of a vertical cliff 25 m above the sea, with speed 20 m/s at 37° above the horizontal (sin 37° = 0.60, cos 37° = 0.80). Air resistance is negligible. Use g = 10 m/s². How long after it is thrown does the ball reach the sea?
Answer and reasoning
A5.0 s A student who uses the launch speed, 20 m/s, as the initial vertical velocity picks this: −25 = 20t − 5t² gives t = 5.0 s. Only the vertical component, 20 sin 37° = 12 m/s, belongs in the vertical equation.
B4.3 s A student who takes the vertical component as 20 cos 37° = 16 m/s picks this: −25 = 16t − 5t² gives t ≈ 4.3 s. The angle is measured from the horizontal, so the vertical component is 20 sin 37° = 12 m/s.
C2.2 s A student who treats the throw like a horizontal launch, with zero initial vertical velocity, picks this: 25 = 5t² gives t ≈ 2.2 s. The ball starts with 12 m/s upward, so it first rises and takes longer to reach the sea.
D3.7 sCorrect Only the vertical motion sets the time. With up positive, vy0 = (20 m/s)(0.60) = 12 m/s and the sea is at y = −25 m: −25 = 12t − 5t². The positive root is t = [12 + √(144 + 500)]/10 ≈ 3.7 s.
Working Up +, origin at launch: vy0 = 20 × 0.60 = 12 m/s. −25 = 12t − 5t² → 5t² − 12t − 25 = 0 → t = [12 + √(144 + 500)]/10 = (12 + 25.4)/10 = 3.7 s.
At t = 0, a particle at the origin of the xy plane is moving with speed v₀ in a direction 37° above the +x axis (sin 37° = 0.60, cos 37° = 0.80). For t ≥ 0, its acceleration is a⃗ = −kt ĵ, where k is a positive constant. At what time t > 0 does the particle return to the x-axis?
Answer and reasoning
A2.2√(v₀/k) A student who takes the y-component of the initial velocity as v₀ cos 37° = 0.80v₀ picks this: 0.80v₀t = kt³/6 gives t = √4.8 √(v₀/k) ≈ 2.2√(v₀/k). The 37° angle is measured from the x-axis, so the y-component is v₀ sin 37°.
B1.9√(v₀/k)Correct The y-motion is independent of the x-motion. With vy0 = v₀ sin 37° = 0.60v₀ and ay = −kt, integrating twice gives y = 0.60v₀t − kt³/6. Setting y = 0 gives t² = 3.6v₀/k, so t = √3.6 √(v₀/k) ≈ 1.9√(v₀/k).
C2.4√(v₀/k) A student who uses the speed v₀ as the initial y-velocity, without resolving it, picks this: v₀t = kt³/6 gives t = √6 √(v₀/k) ≈ 2.4√(v₀/k). Only the component v₀ sin 37° = 0.60v₀ is in the y-direction.
D1.1√(v₀/k) A student who uses y = vy0 t + (1/2)ay t² with the acceleration at the end, ay = −kt, picks this: 0.60v₀t = (1/2)(kt)t² gives t = √1.2 √(v₀/k) ≈ 1.1√(v₀/k). That equation holds only for constant acceleration; here ay changes with time, so vy and y must be found by integration.
Working Only the y-motion matters. vy0 = v₀ sin 37° = 0.60v₀. vy(t) = 0.60v₀ − ∫₀ᵗ kt′ dt′ = 0.60v₀ − kt²/2. y(t) = ∫₀ᵗ vy dt′ = 0.60v₀t − kt³/6. y = 0 with t > 0: t² = 3.6v₀/k, t = √3.6 √(v₀/k) ≈ 1.9√(v₀/k). (x = 0.80v₀t does not affect when y = 0.) Distractors: vy0 = 0.80v₀ → √4.8 ≈ 2.2; vy0 = v₀ → √6 ≈ 2.4; y = 0.60v₀t − (1/2)(kt)t² → √1.2 ≈ 1.1.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
