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FRQ: Horizontally Launched Projectiles — The Stunt-Jump Investigation

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points

▤ 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 horizontally — with no vertical component of velocity — at a speed v_0 that the coordinator sets by choosing the length of the run-up.

Side view of the stunt geometry: a horizontal launch from the upper edge, a 12.0 m drop to the lower roof and an 18.0 m gap between the two edges.

The lower roof extends 40 m beyond its near edge, so overshooting it is not a concern. The stunt's safety review imposes two constraints: the motorcycle must clear the alley and land on the lower roof, and its speed at the instant of landing must not exceed 22 m/s.

Treat the motorcycle and rider as a single object in projectile motion, and take air resistance to be negligible. The motion is two-dimensional, in a single vertical plane. Use the ground as the reference frame, with the launch edge as the origin, the direction of travel as positive x and upward as positive y. Use g = 10 m/s².

✎ Free Response Questions

(a) The coordinator argues that a larger v_0 keeps the motorcycle in the air for longer, and that this extra time in the air is what lets it cross the alley. State whether the coordinator's reasoning is correct, and justify your answer by referring explicitly to the horizontal and vertical components of the motion.

(b) Derive an expression for the minimum launch speed v_0,min at which the motorcycle just reaches the near edge of the lower roof, in terms of the drop height h, the gap d and g. Then calculate its value.

(c) Derive an expression for the motorcycle's speed at landing, v_land, in terms of v_0, g and h. Then calculate the maximum launch speed v_0,max for which the landing speed does not exceed 22 m/s.

(d) Using your results from parts (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 state the margin it leaves against each of the two constraints, or argue that no choice of v_0 is safe. Your argument must explain, in physical terms, why the two constraints push v_0 in opposite directions, and must state the condition on the drop height under which any safe launch speed can exist at all.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

The coordinator's reasoning is incorrect. The horizontal and vertical components of a projectile's motion are independent of each other. The vertical motion starts with v_y0 = 0 for every horizontal launch 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) = √(2 × 12.0 / 10) = 1.55 s

Whether the motorcycle leaves the edge at 8 m/s or at 16 m/s, it reaches the level of the lower roof 1.55 s later. A larger v_0 does increase the horizontal distance covered in that time, because x = v_0 · t with t fixed — 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 time in the air is gained.

Three horizontal launches from the same edge. All three reach the level of the lower roof at t = 1.55 s; only the horizontal distance covered in that time differs. At 8 m/s the motorcycle is still 5.6 m short of the roof edge when it reaches roof level, and falls into the alley.

Scoring (2 points):

▸ 1 point: States that the coordinator is incorrect because the time of flight is independent of v_0 — it is fixed by the drop height and g alone.

▸ 1 point: Justifies the claim through the independence of the components: the vertical motion starts from v_y0 = 0 and is governed only by g and h, while the horizontal motion proceeds at constant velocity, so a larger v_0 increases x = v_0 · t through the speed and not through t.

⚠︎ Common error (partial credit): reaches the correct conclusion but supports it only with “a faster launch does not change the fall”, with no reference to the independence of the components or to v_y0 = 0 — earns 1 of 2 points.

✗ Common error (no credit): agrees 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, which is the misconception the question targets.

Part (b) — Model Answer

With upward positive, the vertical displacement to the lower roof is Δy = −h, the initial vertical velocity is v_y0 = 0 and the acceleration is a_y = −g. Starting from the kinematic equation for displacement under constant acceleration:

Δy = v_y0 · t + ½ · a_y · t²  ⇒  −h = −½ · g · t²  ⇒  t = √(2h/g)

Throughout that time the horizontal motion is at constant velocity (a_x = 0), so the horizontal displacement is x = v_0 · t. To just reach the near edge of the lower roof, x must equal the gap d:

v_0,min · √(2h/g) = d  ⇒  v_0,min = d · √(g / 2h)

Substituting h = 12.0 m, d = 18.0 m and g = 10 m/s²:

t = √(2 × 12.0 / 10) = √2.40 = 1.55 s
v_0,min = 18.0 × √(10 / 24.0) = 18.0 / 1.55 = 11.6 m/s

A launch speed of at least 11.6 m/s is required. Any smaller speed leaves the motorcycle short of the far edge when it has fallen the full 12.0 m, so it drops into the alley.

Scoring (2 points):

▸ 1 point: Begins from an equation-sheet kinematic relation for the vertical motion with v_y0 = 0 (or the equivalent h = ½gt² for the fall time), together with x = v_0 · t for the horizontal motion. This point is awarded for the correct starting relations alone, before any algebra is performed.

▸ 1 point: Obtains the symbolic result v_0,min = d · √(g / 2h) and evaluates it to 11.6 m/s with units.

⚠︎ Common error (partial credit): sets the problem up correctly but makes an arithmetic slip in t or in the final division, or reports the numerical value with no symbolic expression — earns 1 of 2 points, for the starting relations.

✗ Common error (no credit): sets v_y0 = v_0 for a horizontal launch, or substitutes the 18.0 m horizontal gap into the 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, because nothing accelerates the motorcycle horizontally: v_x = v_0. The vertical component at landing follows from the drop, starting from the kinematic equation that links velocity to displacement:

v_y² = v_y0² + 2 · a_y · Δy = 0 + 2(−g)(−h) = 2gh

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

v_land = √(v_x² + v_y²) = √(v_0² + 2gh)

Setting v_land equal to the 22 m/s cap with h = 12.0 m and g = 10 m/s²:

v_0,max = √(v_land² − 2gh) = √(22² − 2 × 10 × 12.0) = √(484 − 240) = √244 = 15.6 m/s

Any launch speed above 15.6 m/s produces a landing speed above the cap. Note that the vertical component at landing, √(2gh) = 15.5 m/s, is the same for every launch speed; only the horizontal component changes. 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: Begins from a fundamental relation for the vertical component — v_y² = v_y0² + 2a_yΔy, or conservation of energy — and states that v_x = v_0 is unchanged in flight. This point is awarded for the correct starting relations alone, before any algebra is performed.

▸ 1 point: Combines the perpendicular components correctly to v_land = √(v_0² + 2gh) and evaluates v_0,max = 15.6 m/s with units (accept 15.6 to 15.7 m/s).

⚠︎ Common error (partial credit): finds v_y = 15.5 m/s correctly but applies the 22 m/s cap to the vertical component alone, concluding that no launch speed is ruled out — earns 1 of 2 points, for the starting relations.

✗ Common error (no credit): adds the components as scalars, v_land = v_x + v_y, or writes v_land = v_0 + g · t — both treat the magnitude of a vector as the arithmetic sum of its components.

Part (d) — Model Answer

From part (b), clearing the alley requires v_0 ≥ 11.6 m/s. From part (c), keeping the landing speed at or below 22 m/s requires v_0 ≤ 15.6 m/s. Because 11.6 < 15.6, the two constraints overlap and a safe window exists:

11.6 m/s ≤ v_0 ≤ 15.6 m/s

Recommend v_0 = 13.5 m/s, close to the middle of that window. At that speed the motorcycle covers x = 13.5 × 1.55 = 20.9 m horizontally, clearing the near edge by 2.9 m, and lands at v_land = √(13.5² + 240) = √422 = 20.5 m/s, 1.5 m/s below the cap. Both margins are comfortable, so a small error in the run-up breaches neither 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 1.55 s sets 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 √(2gh) = 15.5 m/s to give v_land — so the landing-speed cap sets a ceiling on the very quantity the gap forces upward. The trade is not one-for-one: the components combine in quadrature, so adding 1.0 m/s to v_0 near 13.5 m/s raises v_land by only about 0.7 m/s.

A window exists at all only because the fixed vertical contribution √(2gh) sits below the cap. Had the drop been 24.2 m or more, √(2gh) alone would reach 22 m/s, and no launch speed — not even zero — could satisfy the landing constraint.

Landing speed against launch speed. The shaded band is the safe window, from the gap-clearing floor at 11.6 m/s to the landing-speed cap at 15.6 m/s; the marked point is the recommended 13.5 m/s.

Scoring (2 points):

▸ 1 point: Establishes that v_0,min < v_0,max, states the window as an inequality with both bounds and units, recommends a v_0 inside it, and quantifies both margins — clearance in metres and speed buffer in m/s.

▸ 1 point: Explains why the constraints oppose — the fixed fall time makes the gap a floor on v_0, while v_0 is carried unchanged to landing and combines with the fixed v_y = √(2gh) — and states that a window exists only when √(2gh) is below the cap (h < 24.2 m).

⚠︎ Common error (partial credit): names a safe speed and gives the fixed-fall-time argument and the condition on h, but calculates no margins — earns 1 of 2 points, for the explanation.

⚠︎ Common error (partial credit): gives the window, a recommended speed and both margins, but the explanation only restates the numbers, naming neither the fixed fall time nor the condition on h — earns 1 of 2 points, for the window and margins.

✗ Common error (no credit): compares v_0 directly against the 22 m/s cap, or claims each extra 1 m/s of launch speed adds 1 m/s to the landing speed — the first sets a launch speed against a landing speed; the second treats components that add in quadrature as though they added arithmetically