🏍️FRQ: The Stunt-Jump Investigation

Solve the problem with me first, then use the simulation at the end of the lesson to see the concept in action. (Assessments aligned to 2026 AP Physics 1 standards)

📋Scenario

A film stunt coordinator is designing a sequence in which a motorcyclist drives off the edge of a flat-topped warehouse roof and lands on a lower flat-topped warehouse roof. The launch roof is 12.0 m above the lower roof. The motorcycle leaves the launch edge moving purely horizontally (no upward component) at speed v₀, which the coordinator can adjust. The horizontal gap between the two roof edges is 18.0 m. Air resistance is negligible. Take g = 9.8 m/s².

Two safety constraints apply: (1) the motorcycle must clear the gap and land on the lower roof, and (2) at the moment of landing, the speed of the motorcycle must not exceed 22 m/s for the suspension to absorb the impact safely.

(a) State, in a single sentence, the scientific question that constrains the choice of v₀.

(b) Determine the minimum launch speed v₀,min that allows the motorcycle to clear the 18.0 m gap. Show all reasoning.

(c) Determine the maximum launch speed v₀,max that satisfies the landing-speed safety constraint. Show all reasoning.

(d) Based on your answers in (b) and (c), determine whether a safe launch speed exists for this stunt. Construct a complete argument that either (i) recommends a specific safe v₀ to the coordinator, supported by your analysis, or (ii) argues that the geometry of the stunt makes it unsafe regardless of the chosen v₀, supported by your analysis.

📝Answer Key & Scoring Guide

Part (a) — Model Answer

"What range of horizontal launch speeds simultaneously allows the motorcycle to clear an 18.0 m horizontal gap while striking the lower roof at a speed no greater than 22 m/s?"

Scoring (1 point):

✔️1 point: Articulates the question with both constraints clearly stated and linked to v₀.

⚠️Common error: Formulating a question that only addresses the kinematic range (clearing the gap) while completely ignoring the structural constraint (the 22 m/s maximum landing speed). Another frequent mistake is phrasing the response as a factual statement or a procedure rather than a testable scientific question linking the independent variable (v₀) to the dependent constraints.

Part (b) — Model Answer

Time of fall is set by the vertical drop. With launch angle 0 and h = 12.0 m:

t = √(2h/g) = √(2·12.0/9.8) = √(2.449) ≈ 1.565 s

To just clear the 18.0 m gap, the horizontal range R must equal 18.0 m:

R = v₀ · t ⟹ v₀,min = R / t = 18.0 / 1.565 ≈ 11.5 m/s

Scoring (3 points):

✔️1 point: Correctly applies t = √(2h/g) for the horizontal-launch fall time.

✔️1 point: Correctly identifies that v₀,min comes from R = v₀t with R = 18.0 m.

✔️1 point: Correct numerical answer v₀,min ≈ 11.5 m/s with units.

⚠️Common error: Confusing horizontal and vertical kinematic components. Students often incorrectly set the initial vertical velocity (v₀_y) equal to the launch speed (v₀) instead of recognizing that for a strictly horizontal launch, v₀_y = 0 m/s. Another classic mistake is plugging the horizontal gap distance (18.0 m) into a vertical motion equation to solve for time.

Part (c) — Model Answer

The landing speed of a horizontally-launched projectile from height h is, by energy conservation in the absence of air resistance:

v_land = √(v₀² + 2gh)

Setting v_land = 22 m/s and h = 12.0 m, and solving for v₀:

22² = v₀² + 2(9.8)(12.0)

484 = v₀² + 235.2

v₀² = 248.8 ⟹ v₀,max ≈ 15.8 m/s

Scoring (3 points):

✔️1 point: Correctly identifies energy conservation (or equivalent component method) for landing speed in the horizontal-launch case.

✔️1 point: Correct setup with v_land = 22 m/s and h = 12.0 m.

✔️1 point: Correct numerical answer v₀,max ≈ 15.7–15.8 m/s with units.

⚠️Common error: Treating velocity as a scalar rather than a vector. Students frequently assume the 22 m/s limit applies only to the final vertical velocity component, or they attempt to algebraically add the final components (v_land = v_x + v_y) instead of correctly using the Pythagorean theorem (v_land = √(v_x² + v_y²)). Using conservation of energy elegantly avoids this vector trap entirely.

Part (d) — Model Answer

From part (b), the minimum launch speed needed to clear the gap is v₀,min ≈ 11.5 m/s. From part (c), the maximum launch speed that satisfies the landing-speed safety constraint is v₀,max ≈ 15.8 m/s. Because v₀,min < v₀,max, there exists a non-empty range of safe launch speeds: 11.5 m/s ≤ v₀ ≤ 15.8 m/s.

Recommendation: a launch speed near the middle of this range — for example v₀ ≈ 13.5 m/s — provides a safety margin against errors at both ends. At this speed, the motorcycle would clear the gap with horizontal margin (v₀·t − 18.0) = (13.5·1.565 − 18.0) ≈ 3.1 m, and the landing speed √(13.5² + 235.2) ≈ 20.3 m/s leaves a 1.7 m/s buffer below the 22 m/s safety limit.

The argument rests on two physical principles working in tension: the gap-clearing constraint requires a high horizontal speed (since fall time is fixed by the drop height), while the landing-speed constraint caps the launch speed because every additional unit of v₀ adds directly to the landing speed via the Pythagorean combination of vₓ and vᵧ. The geometry is permissive only because the safe-landing window (≈ 4.3 m/s wide) is wider than the gap-clearing minimum demands.

Scoring (3 points):

✔️1 point: Correctly establishes that v₀,min < v₀,max and therefore a safe range exists.

✔️1 point: Recommends a specific v₀ within the safe range with quantitative justification (margin against either constraint).

✔️1 point: Provides a complete physical argument identifying the competing constraints (gap-clearing vs. landing-speed safety) and how they jointly determine the feasible launch-speed window.

✔️Partial credit: Recommendation without margin analysis earns 1 of the 2 recommendation points.

⚠️Common error: Simply picking a number between the minimum and maximum values (e.g., "choose 14 m/s because it is between 11.5 and 15.8") without constructing a quantitative argument to justify the choice. To earn the recommendation points, the student must actually calculate the specific safety margins (the physical buffer distance and the buffer speed) for their chosen v₀ to prove it is a robust and safe recommendation.

🛸FRQ: Drone Off a Cliff

📋Scenario

A drone is flying horizontally at speed 18 m/s at the top of a cliff. The cliff is 45 m above the flat ground below. The drone shuts down and from that moment can be treated as a projectile (no thrust, no drag). Air resistance is negligible. Use g = 9.8 m/s².

All motion is two-dimensional. Take rightward (the direction of the drone's initial motion) as positive x, and upward as positive y. The drone's shutdown position is the origin; the ground is at y = -45 m.

📝Free Response Questions

(a) Predict, without calculation, whether the time the drone takes to fall to the ground will be the same as, greater than, or less than the time a stationary object dropped from rest at the cliff top would take. Justify your prediction by referring to the independence of horizontal and vertical motion. 

(b) Calculate the time it takes the drone to reach the ground after it shuts down.

(c) Calculate the horizontal distance from the base of the cliff at which the drone strikes the ground.

(d) A student claims: 'To get the maximum horizontal distance from the cliff base, the drone should be launched at 45° above horizontal, since 45° always gives maximum range.' Determine whether this claim is correct for this scenario, and explain your reasoning. (You do not need to calculate the optimal angle - you only need to argue whether the 45° claim holds when launch height differs from landing height.) 

🗝️Answer Key & Scoring Guide

Part (a) - Model Answer

The two times are the same. The horizontal and vertical components of motion are independent because the only force acting on the drone is gravity, which acts purely vertically and produces no horizontal acceleration.

The horizontal velocity of 18 m/s does not affect how long the drone takes to fall vertically through 45 m. A stationary object dropped from rest at the same height has the same vertical motion (same initial vertical velocity = 0, same vertical acceleration = -g) and therefore the same fall time.

Diagram below: Stationary object and drone land at the same time

Scoring (3 points):

✔️1 point: Correct prediction that the times are equal.

✔️1 point: Justification invokes independence of horizontal and vertical motion.

✔️1 point: Justification correctly attributes independence to gravity acting purely vertically (no horizontal component of force, so no horizontal acceleration).

⚠️Common error: predicting equal times but justifying by 'no air resistance' rather than perpendicular-component independence - earns 1 of 3 points.

Part (b) - Model Answer

Apply Δy = v₀ᵧ t + ½ a_y t² with v₀ᵧ = 0 (horizontal launch), Δy = -45 m, a_y = -9.8 m/s²:

-45 = 0 + ½ × (-9.8) × t²

t ≈ 3.03 s

The drone takes approximately 3.0 s to reach the ground.

Scoring (2 points):

✔️1 point: Correct kinematic equation and substitution with v₀ᵧ = 0 and consistent signs.

✔️1 point: Correct numerical answer t ≈ 3.0 s (accept 2.95-3.05 s).

⚠️Common error: using v₀ = 18 m/s as v₀ᵧ - earns 0 points.

Part (c) - Model Answer

Horizontal motion has zero acceleration; horizontal velocity is constant at v_x = 18 m/s. Using x = v_x · t:

x = 18 × 3.03 ≈ 54.5 m

The drone strikes the ground approximately 54.5 m from the base of the cliff.

Scoring (2 points):

✔️1 point: Correct identification that horizontal velocity is constant and use of x = v_x · t.

✔️1 point: Correct numerical answer x ≈ 54-55 m, using the time from part (b).

✔️Earn-back: full credit awarded for consistent use of incorrect time from (b) if method is correct.

Part (d) - Model Answer

The claim is incorrect for this scenario. The result that 45° gives maximum range comes from the formula R = v₀² sin(2θ₀)/g, which was derived assuming the projectile lands at the same height from which it was launched.

For a launch from a cliff (launch height higher than landing height), the projectile spends additional time in the air after passing the launch level, and during that extra time the horizontal velocity component continues to carry the projectile forward.

A lower launch angle - which gives a larger horizontal velocity component v₀ cosθ₀ - takes advantage of this extra fall time and produces a larger horizontal range than 45° would.

Therefore, for an elevated launch the optimal launch angle is less than 45°, and the maximum horizontal distance from the cliff base is greater than v₀²/g.

Scoring (3 points):

✔️1 point: Identifies the claim as incorrect for this scenario.

✔️1 point: Correctly explains that the 45° result depends on the same-height assumption built into the range formula.

✔️1 point: Provides physical reasoning that a lower angle increases the horizontal velocity component, which - combined with extra fall time from the cliff - gives a larger range.

⚠️Common error: agreeing with the student because '45° always gives maximum range' - earns 0 points (this is the precise misconception the question targets).