FRQ: Horizontally Launched Projectiles — The Stunt-Jump Investigation

Assessments aligned to 2026 AP Physics 1 standards

Question type: Qualitative/Quantitative Translation (QQT) 

 ▤ Scenario

A film stunt coordinator is designing a sequence in which a motorcyclist rides off the edge of a flat warehouse roof and lands on the roof of a lower warehouse across an alley. The launch edge is 12.0 m above the lower roof, and the horizontal gap between the two roof edges is 18.0 m. The motorcycle leaves the edge moving purely horizontally — no upward component — at a speed v_0 the coordinator sets by choosing the run-up length.

Figure 1 — Side view of the stunt geometry: a 12.0 m drop across an 18.0 m gap.

The lower roof extends 40 m beyond its near edge, so overshooting the far side is not a concern. Two safety constraints apply:

1. The motorcycle must clear the alley and land on the lower roof.

2. At landing, its speed must not exceed 22 m/s, so the suspension can absorb the impact.

Treat the motorcycle and rider as a single object in projectile motion, and neglect air resistance. Motion is two-dimensional in one vertical plane. Take the launch edge as the origin, the direction of travel as positive x, upward as positive y, and the ground as the reference frame. Use g = 9.8 m/s².

✎ Free Response Questions

(a) The coordinator argues that choosing a larger v_0 will keep the motorcycle in the air longer, and that the extra hang time is what lets it cross the alley. State whether the coordinator is correct, and justify your answer by referring explicitly to the horizontal and vertical components of the motion. (2 points)

(b) Determine the minimum launch speed v_0,min for which the motorcycle reaches the near edge of the lower roof. Show all reasoning. (3 points)

(c) Determine the maximum launch speed v_0,max for which the landing speed does not exceed the 22 m/s limit. Show all reasoning. (2 points)

(d) Using your answers to (b) and (c), determine whether a safe launch speed exists for this stunt. In a paragraph-length response, either recommend a specific value of v_0 and justify it by stating the margin it leaves against each of the two constraints, or argue that the geometry makes the stunt unsafe for every choice of v_0. Your argument must explain, in physical terms, why the two constraints push v_0 in opposite directions. (3 points)

❖ Answer Key & Scoring Guide

▸ earns credit    ⚠︎ common error, partial credit    ✗ common error, no credit

Part (a) — Model Answer

The coordinator is incorrect. For a projectile launched horizontally the vertical and horizontal motions are independent. The vertical motion begins from rest in that direction (v_0y = 0) and is governed only by the free-fall acceleration and the drop height, so the time to fall 12.0 m is the same for every launch speed:

t = √(2h/g)

That time is fixed at about 1.56 s whether the motorcycle leaves at 8 m/s or at 16 m/s. A larger v_0 does increase the horizontal distance covered, because x = v_0 · t, but it does so by increasing the horizontal speed — not by extending the time of flight. The coordinator has the right conclusion for the wrong reason: more speed does help clear the alley, but no extra hang time is purchased.

Figure 2 — Three horizontal launches from the same edge. The trajectories differ in horizontal reach but all three arrive at the lower roof at t = 1.56 s, showing that time of flight is independent of v_0.

Scoring (2 points):

▸ 1 point: States that the coordinator is incorrect AND that the time of flight is independent of v_0.

▸ 1 point: Justifies the claim by the independence of the components — vertical motion set by g and h alone with v_0y = 0, horizontal motion at constant velocity.

⚠︎ Common error (partial credit): Correct conclusion supported only by “larger speed does not change the fall” with no reference to component independence or to v_0y = 0 — earns 1 of 2 points.

✗ Common error (no credit): Agreeing with the coordinator on the grounds that a faster projectile “stays up longer” — this treats the horizontal speed as though it feeds the vertical motion.

Part (b) — Model Answer

The fall time is set entirely by the vertical drop. Taking the downward displacement to have magnitude h = 12.0 m and starting from v_0y = 0:

Δy = v_0y · t + ½ · a · t²  ⇒  h = ½ · g · t²  ⇒  t = √(2h/g)

t = √(2 × 12.0 / 9.8) = √2.449 = 1.565 s

Throughout that time the horizontal motion is at constant velocity, so the horizontal displacement is x = v_0 · t. To just reach the near edge of the lower roof, x must equal the 18.0 m gap:

v_0,min = x / t = 18.0 / 1.565 = 11.5 m/s

A launch speed of at least about 11.5 m/s is required. Any smaller speed puts the motorcycle into the alley before it has fallen the full 12.0 m.

Scoring (3 points):

▸ 1 point: Begins the derivation from a fundamental kinematic relation on the official equation sheet — Δy = v_0y · t + ½ · a · t² with v_0y = 0 — rather than quoting t = √(2h/g) as a memorised result.

▸ 1 point: Correct fall time, t ≈ 1.56 s.

▸ 1 point: Correct v_0,min ≈ 11.5 m/s, with units, obtained from x = v_0 · t with x = 18.0 m.

⚠︎ Common error (partial credit): Correct physical setup with an arithmetic slip in t or in the final division — earns 2 of 3 points, provided the component reasoning is sound.

⚠︎ Common error (partial credit): Rounding t to 1.6 s and reporting v_0,min = 11.3 m/s — earns 3 of 3 if the method is shown, since the rounding is stated, but flag it as premature rounding.

✗ Common error (no credit): Setting v_0y = v_0 for a horizontal launch, or substituting the 18.0 m horizontal gap into a vertical kinematic equation to find t — both destroy the independence of the components on which the whole solution rests.

Part (c) — Model Answer

The landing speed is the magnitude of the velocity vector at impact, not either component on its own. The horizontal component is unchanged throughout the flight, so v_x = v_0. The vertical component follows from the drop:

v_y² = v_0y² + 2 · a · Δy  ⇒  v_y² = 2 · g · h

The two components are perpendicular, so they combine by the Pythagorean theorem:

v_land = √(v_x² + v_y²) = √(v_0² + 2 · g · h)

Setting v_land equal to the 22 m/s limit with h = 12.0 m:

22² = v_0² + 2(9.8)(12.0)

484 = v_0² + 235.2  ⇒  v_0² = 248.8  ⇒  v_0,max = 15.8 m/s

Any launch speed above about 15.8 m/s brings the motorcycle down faster than the suspension can absorb. The same relation follows from conservation of energy, ½mv_land² = ½mv_0² + mgh, which gives v_land² = v_0² + 2gh directly — either route earns full credit.

Scoring (2 points):

▸ 1 point: Starts from a fundamental principle — either v_y² = v_0y² + 2aΔy combined with vector addition of perpendicular components, or conservation of energy — to reach v_land² = v_0² + 2gh.

▸ 1 point: Correct v_0,max with units; accept 15.7 m/s to 15.8 m/s.

⚠︎ Common error (partial credit): Correct relation written but the 22 m/s limit applied to v_y alone, giving no solution or a nonsensical v_0 — earns 1 of 2 points if the Pythagorean combination is otherwise stated correctly.

✗ Common error (no credit): Adding the components algebraically as v_land = v_x + v_y — this treats the magnitude of a vector as the scalar sum of its components.

✗ Common error (no credit): Using v_land = v_0 + g · t. This is a vertical-component relation applied to a speed, and it mixes a horizontal quantity into a one-dimensional kinematic equation.

Part (d) — Model Answer

From part (b), clearing the alley requires v_0 ≥ 11.5 m/s. From part (c), keeping the landing speed at or below 22 m/s requires v_0 ≤ 15.8 m/s. Because 11.5 < 15.8, the two constraints overlap and a non-empty safe window exists:

11.5 m/s ≤ v_0 ≤ 15.8 m/s

Recommend v_0 = 13.5 m/s, near the middle of that window. At that speed the motorcycle covers x = 13.5 × 1.565 = 21.1 m horizontally, clearing the near edge by 3.1 m, and lands at v_land = √(13.5² + 235.2) = 20.4 m/s, which sits 1.6 m/s below the 22 m/s ceiling. Both margins are comfortable, so a small error in the run-up does not breach either constraint.

Physically the two constraints pull in opposite directions because the fall time is fixed by the drop height and cannot be altered by the rider. Crossing a fixed 18.0 m gap within a fixed 1.565 s therefore imposes a floor on the horizontal speed. But that same horizontal speed survives untouched to the moment of landing, where it combines with the fixed vertical impact speed v_y = √(2gh) = 15.3 m/s to set v_land. Every metre per second of horizontal speed bought to widen the clearance is also carried into the impact, so the safety limit imposes a ceiling. The stunt is feasible only because the fixed vertical contribution of 15.3 m/s leaves room under the 22 m/s ceiling for a horizontal component as large as 15.8 m/s — comfortably more than the 11.5 m/s the gap demands.

One refinement worth noting: the two speeds do not trade one-for-one. Because the components combine in quadrature rather than by simple addition, adding 1.0 m/s to v_0 near 13.5 m/s raises v_land by only about 0.7 m/s. The safety margin therefore erodes more slowly than the launch speed grows, which is what makes a window of this width available at all.

Figure 3 — Landing speed as a function of launch speed. The shaded band is the safe window bounded by the gap-clearing floor at 11.5 m/s and the 22 m/s landing-speed ceiling at 15.8 m/s; the marked point is the recommended v_0 = 13.5 m/s.

Scoring (3 points):

▸ 1 point: Establishes that v_0,min < v_0,max and states the safe window as an inequality carrying both bounds and units.

▸ 1 point: Recommends a specific v_0 inside the window AND quantifies the margin against both constraints — clearance in metres and speed buffer in m/s.

▸ 1 point: Paragraph-length argument explaining why the constraints oppose: fall time is fixed by h, so the gap sets a floor on v_0, while v_0 carries through to landing and combines with the fixed v_y.

⚠︎ Common error (partial credit): Names a safe speed and asserts it lies between 11.5 and 15.8 m/s, but calculates no margins — earns 1 of 3 points, for the window only.

⚠︎ Common error (partial credit): Correct window and margins, but the explanation restates the numbers without naming the fixed fall time or the perpendicular combination — earns 2 of 3 points.

✗ Common error (no credit): Claiming each extra 1 m/s of launch speed adds 1 m/s to the landing speed — the components add in quadrature, so the real increase near v_0 = 13.5 m/s is about 0.7 m/s.

✗ Common error (no credit): Comparing v_0 directly against the 22 m/s limit — this sets a launch speed against a landing speed.

Total: 10 points — (a) 2 | (b) 3 | (c) 2 | (d) 3

Note on gravity: g = 9.8 m/s² here, matching the official 2026 AP Physics 1 reference sheet and the source scenario — a deliberate departure from the g = 10 m/s² house convention

FRQ: Projectile Motion from an Elevated Launch — Drone Off a Cliff

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  10 points

▤ Scenario

A survey drone is flying horizontally at a constant speed of 18 m/s along the top edge of a coastal cliff, photographing a seabird nesting site. The cliff top is 45 m above the flat beach below. At one instant the battery cuts out: the rotors stop, thrust ends, and from that moment the drone can be treated as a projectile.

Air resistance is negligible. Use g = 10 m/s².

Take the drone’s shutdown position as the origin, with rightward — the direction of the drone’s initial motion — as positive x and upward as positive y. The beach lies at y = −45 m. All motion is two-dimensional and confined to a single vertical plane, and the ground is the reference frame throughout.

Figure 1 — The drone at the instant of shutdown: launch height 45 m above the beach, initial velocity 18 m/s directed horizontally, origin at the shutdown position.

✎ Free Response Questions

(a) Predict, without calculation, whether the time the drone takes to reach the beach is greater than, less than, or the same as the time taken by a small stone released from rest at the cliff edge at the same instant. Justify your prediction in terms of the independence of the horizontal and vertical components of the motion.  (3 points)

(b) Calculate the time it takes the drone to reach the beach after shutdown.  (2 points)

(c) Calculate the horizontal distance from the base of the cliff at which the drone lands.  (2 points)

(d) A student looks at your answer to part (c) and claims:

“The drone would have landed at its greatest possible distance from the base of the cliff if it had been launched at 45° above the horizontal at the same speed, because the range of a projectile is R = v_0²·sin(2θ)/g, which is largest when θ = 45°.”

Write a paragraph-length coherent argument evaluating this claim. Your argument must (i) state whether the claim is correct for this scenario, (ii) identify the condition assumed in deriving R = v_0²·sin(2θ)/g that this scenario violates, and (iii) use your numerical answer from part (c) together with the value of v_0²/g for this drone as quantitative evidence. You do not need to calculate the optimal launch angle.  (3 points)

❖ Answer Key & Scoring Guide

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part (a) — Model Answer

The two times are the same.

After shutdown the only force acting on the drone is the gravitational force, and that force is directed vertically downward. It has no horizontal component, so a_x = 0 and nothing about the horizontal motion can alter the vertical motion. The horizontal and vertical components of projectile motion are independent of one another.

Vertically, the drone and the stone begin under identical conditions. The drone’s 18 m/s is entirely horizontal, so v_0y = 0 for both objects; both fall through the same Δy = −45 m; and both have a_y = −g. Identical vertical initial conditions under an identical vertical acceleration produce identical vertical motion, so the two objects reach the beach at the same instant. The 18 m/s determines how far downrange the drone is when it lands — not when it lands.

Figure 2 — The drone (teal) and a stone released from rest at the cliff edge (navy). The dashed amber lines mark equal heights at t = 1.0 s, 2.0 s and 3.0 s; the vertical motions are identical and both objects land at t = 3.0 s.

Scoring (3 points):

▸ 1 point: States that the two times are the same.

▸ 1 point: Justification invokes the independence of the horizontal and vertical components of the motion.

▸ 1 point: Attributes that independence to the gravitational force acting vertically only — no horizontal force component, therefore a_x = 0 — or equivalently establishes that both objects share v_0y = 0 and a_y = −g over the same Δy.

⚠︎ Common error (partial credit): Correct prediction of equal times justified only by “air resistance is negligible” — earns 1 of 3 points. Negligible drag is a stated condition of the problem, not the reason the perpendicular components are independent.

✗ Common error (no credit): Predicting the drone takes longer because “it also has to travel horizontally” — earns 0 points. This treats horizontal displacement as competing with the vertical motion for the same time.

Part (b) — Model Answer

Work in the vertical direction with the constant-acceleration kinematic relation. Because the launch is horizontal, v_0y = 0.

Δy = v_0y·t + ½·a_y·t²

Substituting Δy = −45 m, v_0y = 0 and a_y = −10 m/s²:

−45 = 0 + ½·(−10)·t²

t² = 9.0 ⇒ t = 3.0 s

The drone reaches the beach 3.0 s after shutdown.

Scoring (2 points):

▸ 1 point: Selects the correct vertical kinematic relation with v_0y = 0, and applies signs consistently to Δy and a_y.

▸ 1 point: Correct value t = 3.0 s (accept 2.9–3.1 s).

⚠︎ Common error (partial credit): Omitting the factor ½, giving t ≈ 2.1 s — earns 1 of 2 points for correct equation selection and substitution.

✗ Common error (no credit): Substituting 18 m/s for v_0y — earns 0 points. The launch velocity is entirely horizontal, so its vertical component is zero.

Part (c) — Model Answer

The gravitational force is vertical, so a_x = 0 and the horizontal velocity component keeps its initial value for the whole flight:

v_x = 18 m/s (constant)

With zero horizontal acceleration the horizontal displacement is simply speed × time, using the flight time from part (b):

x = v_x·t = (18)·(3.0) = 54 m

The drone lands 54 m from the base of the cliff, measured along the beach in the +x direction.

Figure 3 — The drone’s trajectory. The 45 m vertical drop fixes the flight time at 3.0 s; the constant horizontal velocity then fixes the landing point at x = 54 m.

Scoring (2 points):

▸ 1 point: States or uses that the horizontal velocity is constant because a_x = 0, and applies x = v_x·t.

▸ 1 point: Correct value x = 54 m (accept 53–55 m).

▸ Full credit is awarded for a correct method carried out consistently with an incorrect time imported from part (b).

⚠︎ Common error (partial credit): Using the drone’s speed at impact, √(18² + 30²) ≈ 35 m/s, in place of v_x — earns 1 of 2 points if x = v·t is otherwise correctly applied.

✗ Common error (no credit): Applying a horizontal acceleration — for example writing x = v_x·t + ½·a_x·t² with a_x ≠ 0 — earns 0 points. There is no horizontal component of force, so there is no horizontal acceleration.

Part (d) — Model Answer

The claim is incorrect for this scenario.

The result R = v_0²·sin(2θ)/g is derived on the assumption that the projectile returns to the same vertical height from which it was launched, so that the net vertical displacement over the flight is zero. That condition is not satisfied here. The drone is launched from the cliff top and lands 45 m below its launch height, so the derivation — and with it the “maximum at 45°” conclusion drawn from it — does not apply.

The numbers make the failure explicit. Because sin(2θ) ≤ 1, the largest range that formula permits at any launch angle is

v_0²/g = (18)²/(10) = 32.4 m

Yet part (c) shows the drone lands 54 m from the base of the cliff after a purely horizontal launch — a launch for which the same formula predicts R = 0. A distance of 54 m measured against a formula ceiling of 32.4 m is direct quantitative evidence that this relation is not governing the situation.

Physically, launching from an elevated position means the projectile keeps falling after it has passed its launch level, and that extra time aloft is time during which the constant horizontal component v_0·cosθ continues to carry it forward. Lowering the launch angle below 45° increases v_0·cosθ, and for an elevated launch that gain in horizontal speed outweighs the lost hang time down to some angle below 45°. The maximum horizontal distance is therefore reached at a launch angle less than 45°, not at 45°.

Note on the near miss: 45° is not the worst choice here — it does beat a horizontal launch. What fails is the claim that 45° is the maximum. For this cliff and this launch speed the optimum lies between 0° and 45°, so a student who “corrects” the claim by asserting that a horizontal launch is best has overshot in the other direction.

Figure 4 — Three launches at the same speed v_0 = 18 m/s from the same 45 m cliff. The 45° launch (teal, 58 m) beats the horizontal launch (navy, 54 m), but neither is optimal: the greatest distance comes from a launch angle below 45° (amber, ≈ 27°, 63 m).

Scoring (3 points):

▸ 1 point: States that the claim is incorrect — 45° does not give the maximum horizontal distance in this scenario.

▸ 1 point: Identifies the equal launch-and-landing-height condition (zero net vertical displacement) assumed in deriving R = v_0²·sin(2θ)/g, and states that the elevated launch violates it.

▸ 1 point: Supplies quantitative support — computes v_0²/g = 32.4 m as the formula’s largest possible range and contrasts it with the 54 m obtained in part (c) — and/or gives the physical argument that a lower angle increases v_0·cosθ while the extra fall from the cliff preserves flight time, placing the optimum below 45°.

⚠︎ Common error (partial credit): Concluding “incorrect” and reasoning correctly about the height difference, but offering no quantitative evidence from part (c) or from v_0²/g — earns 2 of 3 points. A QQT argument must connect the qualitative claim to the data.

⚠︎ Common error (partial credit): Rejecting 45° correctly but then asserting that the horizontal launch gives the maximum distance — earns 2 of 3 points. The conclusion about 45° is right, but 0° is not optimal either.

✗ Common error (no credit): Agreeing with the student because “45° always gives maximum range” — earns 0 points. This is precisely the overgeneralisation the question targets: a valid result applied outside the condition under which it was