FRQ : The Two Ball Race

Assessments aligned to 2026 AP Physics 1 standards

Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers

Scenario:

Two students stand on a tall bridge and simultaneously release two balls, A and B, as described below.

🔸Ball A: Released from rest from a height 2H above the ground

🔸Ball B: Thrown straight down at speed v₀ from a height H above the ground

Take the positive y-direction as upward and the ground as y = 0. Air resistance is negligible. Use g for the magnitude of the acceleration due to gravity.

For numerical calculations, use H = 20 m, v₀ = 10 m/s, and g = 10 m/s².

Questions

(a) Starting with a kinematic equation from the reference information, derive an expression for the time t_A it takes Ball A to reach the ground. Express your answer in terms of H and g only. Begin your derivation by writing a fundamental physics equation.

(b) Starting with a kinematic equation from the reference information, derive an expression for the time t_B it takes Ball B to reach the ground. Express your answer in terms of H, g, and v₀.

(c) Using the given numerical values:

🔹Calculate the numerical value of t_A.

🔹Calculate the numerical value of t_B.

State which ball hits the ground first and explain whether this result is physically expected.

(d) After the experiment, a student makes the following claim:

"Ball A must always hit first regardless of v₀, because it falls twice as far and therefore gains more speed from gravity."

In a clear, coherent, paragraph-length response that may reference your expressions from parts (a) and (b):

👉State whether you agree or disagree with the student's claim.

👉Justify your answer by explaining what physical factors determine which ball hits first, and why the student's reasoning is flawed or correct.

Answer Key & The Breakdown

✅ (a) Starting with: y = y₀ + v₀t + ½at²

Substituting y = 0, y₀ = 2H, v₀ = 0, a = −g:

0 = 2H + 0 + ½(−g)t²

½gt² = 2H

t² = 4H / g

t_A = 2√(H/g)

Total for part (a): 2 points

✔️For selecting an appropriate kinematic equation and substituting the correct initial conditions for Ball A: y₀ = 2H, v₀ = 0, a = −g, y = 0. [1 point]

✔️For correctly solving to obtain t_A = √(4H/g) = 2√(H/g). [1 point]

✅ (b) Starting with: y = y₀ + v₀t + ½at²

Substituting y = 0, y₀ = H, v₀ = −v₀ (downward), a = −g:

0 = H + (−v₀)t + ½(−g)t²

0 = H − v₀t − ½gt²

Rearranging: ½gt² + v₀t − H = 0

Using the quadratic formula with A = ½g, B = v₀, C = −H:

t = [−v₀ ± √(v₀² + 2gH)] / g

Since t must be positive: t_B = [−v₀ + √(v₀² + 2gH)] / g

Total for part (b): 2 points

✔️For selecting an appropriate kinematic equation and substituting the correct initial conditions for Ball B: y₀ = H, v₀ = −v₀ (negative because thrown downward), a = −g, y = 0. [1 point]

✔️For correctly rearranging to a quadratic equation and applying the quadratic formula to obtain: t_B = [-v₀ + √(v₀² + 2gH)] / g (taking the positive root). [1 point]

✅ (c) Time t_A and t_B

t_A = 2√(H/g) = 2√(20/10) ≈ 2.83 s

t_B = [−10 + √(100 + 2(10)(20))] / 10 ≈ 1.24 s

Ball B hits the ground first (t_B ≈ 1.24 s < t_A ≈ 2.83 s). Although Ball B only falls half the distance, it has an initial downward velocity of 10 m/s, which gives it a significant head start. Ball A starts from rest and must spend time accelerating from zero speed. For these particular values, Ball B's velocity advantage more than compensates for the shorter fall distance.

Total for part (c): 3 points

✔️For correctly calculating t_A = 2√(20/10) = 2√2 ≈ 2.83 s. [1 point]

✔️For correctly calculating t_B = [−10 + √(100 + 400)] / 10 ≈ 1.24 s. [1 point]

✔️For correctly stating that Ball B hits the ground first and providing a physical reason [1 point]

✅ (d) I disagree with the student's claim. The student's reasoning is flawed because they conflate final speed with time of arrival. While it is true that Ball A falls farther and therefore reaches a greater speed at impact, this does not mean it arrives at the ground sooner. The time of arrival depends on the entire history of the motion — starting velocity, distance traveled, and acceleration — not just the final speed.

Ball A starts from rest (v₀ = 0) at height 2H and must accelerate from zero. Its time is t_A = 2√(H/g), which depends only on H and g and is completely independent of any initial velocity. Ball B, on the other hand, begins with an initial downward speed v₀ that gives it an immediate advantage: it is covering distance from the very first instant, while Ball A is still nearly stationary.

The numerical results from part (c) confirm this: with H = 20 m and v₀ = 10 m/s, Ball B reaches the ground in about 1.24 s while Ball A takes about 2.83 s. Moreover, by examining the expression t_B = [−v₀ + √(v₀² + 2gH)] / g, we can see that as v₀ increases, t_B decreases — meaning Ball B can be made to arrive arbitrarily quickly by increasing v₀. There is no value of v₀ for which Ball A arrives first;

Ball B always wins this particular race for any v₀ > 0, because Ball A's arrival time is fixed.

Total for part (d): 3 points

✔️For a correct claim that disagrees with the student — Ball A does not always hit first. [1 point]

✔️For identifying the flaw in the student's reasoning [1 point]

✔️For a correct justification using the derived expressions or physical reasoning [1 point]