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FRQ: Free-Fall Acceleration from Fall-Time Data — The Two-Level Bridge

Assessments aligned to 2026 AP Physics 1 standards

Experimental Design and Analysis (LAB)  |  CHALLENGE  |  10 points  |  25 min

▤ Scenario

Two students take part in a demonstration on a two-level bridge that spans a dry riverbed. Student 1 stands on the upper deck and holds Ball A at a height 2H above the riverbed. Student 2 stands on the lower deck, directly below, and holds Ball B at a height H above the riverbed. At the same instant, Student 1 releases Ball A from rest and Student 2 throws Ball B straight down with speed v_0. In the demonstration H = 20 m and v_0 = 10 m/s.

Figure 1 — The two-level bridge at the instant of release.

The demonstration leaves the two students arguing about what actually sets the time a ball takes to reach the riverbed, so they decide to measure the free-fall acceleration g on the bridge for themselves. A vertical support column runs the full height from the riverbed to the upper deck, marked at 1.0 m intervals, and a release bracket can be clamped to it at any marked height from 2.0 m to 40.0 m. The students have a measuring tape, stopwatches, an electronic timer that can be started and stopped by a pair of photogates, and a supply of identical steel balls.

All motion is one-dimensional and vertical. Heights above the riverbed are stated as magnitudes, and h denotes both a release height and the distance a ball falls from it. Take the downward direction as positive, so that a distance fallen, a downward speed and the free-fall acceleration g are all positive quantities. Air resistance is negligible.

✎ Free Response Questions

The following information applies to parts A and B.

The students plan an experiment in which a ball is released from rest from several different heights above the riverbed.

Part A

Indicate the quantities that could be measured in this experiment that would allow the students to determine g using a linear graph. Then briefly describe a method for reducing the experimental uncertainty in those measurements, and state the source of uncertainty that the method addresses.

Part B

Indicate what quantities should be graphed on the horizontal axis and on the vertical axis so that the resulting graph is linear. Then briefly describe how g could be determined from a feature of that graph.

The following information applies to parts C and D.

A second group carries out a related experiment on the same bridge. A spring-loaded launcher is fixed to the release bracket, so that at each clamped height it projects a ball vertically downward with the same launch speed. That launch speed is not measured, and it is not v_0. The group clamps the bracket at five different heights h above the riverbed and records the time t for the ball to reach the riverbed. Because the bracket sits at a mark on the column, the uncertainty in each height is negligible compared with the uncertainty in the timing; the timer reads to the nearest millisecond, and repeated trials at the same height scatter by a few milliseconds. The data are shown in Figure 2.

Figure 2 — Release height and time of fall for the launcher trials.

Part C

i.    Calculate the values of a quantity, obtained from the data in Figure 2, that will produce a linear graph when it is plotted on the vertical axis against t on the horizontal axis. Label the vertical axis of the grid in Figure 3 with that quantity, including units.

ii.   Mark a scale on the vertical axis of Figure 3, plot the data, and draw a best-fit line.

Figure 3 — Grid for the graph called for in part C.

Part D

Using the best-fit line, calculate an experimental value for g.

❖ Answer Key & Scoring Guide

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

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

Measure the release height h of the ball above the riverbed, and the time t from release until the ball reaches the riverbed. Both are needed, and nothing else is.

The larger uncertainty is in t. A hand-operated stopwatch carries a human reaction time of order 0.2 s at the start and again at the stop, which is a sizeable fraction of a fall lasting a couple of seconds. Start and stop the electronic timer with the two photogates instead — one level with the release point and one level with the riverbed surface — so that the ball itself opens and closes the timing gate and reaction time is removed altogether.

A ball released from rest carries no launch speed into the problem, so the release height and the fall time are between them enough.

Scoring (2 points):

▸ A1 — 1 point: indicates a set of measurable quantities from which g can be obtained from a linear graph. The release height h above the riverbed together with the corresponding time of fall t is the expected pair; the release height together with the speed at which the ball reaches the riverbed also earns the point.

▸ A2 — 1 point: describes a workable method of reducing the experimental uncertainty in one of the quantities the student has named that the experiment actually requires, and names the source of uncertainty that method addresses. This point is awarded consistent with the quantities named in A1, whether or not A1 was earned.

Scoring Note: any method that genuinely reduces a named source of uncertainty earns A2 — photogates and an electronic timer in place of a hand-operated stopwatch (reaction time), repeated trials averaged at each height (random scatter), or a taut vertical tape read to a fixed point on the ball (parallax). A method is not required for both quantities.

⚠︎ Common error (partial credit): names the height and the fall time correctly, but offers only “measure carefully” or “take several readings” with no source of uncertainty identified — earns A1 only, 1 of 2 points.

✗ Common error (no credit): gives the release height and the ball’s mass as the two quantities to be measured, and describes a method for weighing the ball more precisely. With air resistance negligible the fall time does not depend on mass, so no graph built from mass has a slope from which g follows, and a method addressing a quantity the experiment does not require earns nothing either.

Part B — Model Answer

Plot h on the vertical axis against t² on the horizontal axis.

For a ball released from rest, the kinematic relation for position under constant acceleration reduces to

h = ½·g·t²

which has the form y = m·x with y = h, x = t² and m = ½·g. The graph is therefore a straight line through the origin, and

g = 2 × (slope of the line)

Rearranged, the same relation gives the fall time as t = √(2h/g). Applied to the demonstration drop, where Ball A falls 2H from rest, that is t_A = 2√(H/g) — an expression in which the launch speed does not appear at all.

No launch speed appears anywhere in h = ½·g·t², which is why the graph for a ball released from rest passes through the origin.

Figure 4 — Height against the square of the fall time, for a ball released from rest.

Scoring (2 points):

▸ B1 — 1 point: indicates a pair of axes that linearises the relationship — h on the vertical axis against t² on the horizontal, or t² on the vertical against h on the horizontal. Any other pair of axes that is linear for the quantities the student named in part A also earns the point; for the height-and-landing-speed pair this is v² against h.

▸ B2 — 1 point: states how g follows from a feature of that graph — g is twice the slope for h against t², 2 divided by the slope for t² against h, half the slope for v² against h. This point is awarded consistent with the axes the student named in B1, whether or not B1 was earned.

⚠︎ Common error (partial credit): correct axes, but the relation is quoted as g = slope, losing the factor of 2 that the ½ in h = ½·g·t² carries — earns B1 only, 1 of 2 points.

✗ Common error (no credit): proposes plotting h against t and states that g is the slope of that graph. For a ball released from rest h against t is a parabola, so it has no single slope at all, and naming the slope of a curve as g is not a relation any feature of that graph supports.

Part C (i) — Model Answer

For a ball projected downward with launch speed u from a height h, the same kinematic relation gives

h = u·t + ½·g·t²

Dividing every term by t:

h/t = u + (g/2)·t

which has the form y = m·x + c with y = h/t, x = t, m = g/2 and c = u. So h/t plotted against t is a straight line. Calculating h/t for each trial, to three significant figures:

8.91     10.6     13.5     15.6     17.5     (m/s)

Label the vertical axis h/t (m/s).

Solved for t instead, the same equation gives the closed form t = [−u + √(u² + 2gh)]/g. Written for the demonstration throw, in which Ball B is released downward at v_0 from a height H, that is t_B = [−v_0 + √(v_0² + 2gH)]/g. Ball B reaches the riverbed ahead of Ball A both because it starts closer and because it starts already moving.

The launch speed does not vanish from the analysis, but survives as the constant term.

Scoring (2 points):

▸ C1 — 1 point: identifies h/t — the height divided by the fall time — as a quantity that produces a linear graph against t, and calculates its five values: 8.91, 10.6, 13.5, 15.6 and 17.5 m/s, or the same quotients carried to four figures. Any value within 1% of the correct quotient earns credit, and one slip among the five does not cost the point. This is the first judgment in the part and carries no consistency credit.

▸ C2 — 1 point: labels the vertical axis of Figure 3 with that quantity and its unit — h/t in m/s. This point is awarded consistent with the quantity the student named in C1 and with that quantity’s own unit, whether or not C1 was earned.

⚠︎ Common error (partial credit): identifies h/t and calculates the five values correctly, but labels the vertical axis with no unit, or with the unit of h rather than of h/t — earns C1 only, 1 of 2 points.

⚠︎ Common error (partial credit): plots h itself against t, on the grounds that h and t are what the group recorded, and labels the axis h (m). For this motion h against t is a curve, so C1 is not earned; the axis is correctly labelled for the quantity chosen, so C2 is — 1 of 2 points.

Part C (ii) — Model Answer

A scale of 0 to 18 m/s, one grid division per 1.0 m/s, places all five values on the grid and leaves the vertical intercept visible. The five points lie close to a straight line, which is drawn as a single straight edge passing among them with points falling on both sides of it — not through any one point in particular, and not forced through the origin.

Figure 5 — The completed graph, with the slope taken between two points on the drawn line.

Scoring (2 points):

▸ C3 — 1 point: plots all five of the student’s values to within half a grid division of their positions on the scale the student has marked. Awarded consistent with the values from part C (i), provided the quantity plotted is one that produces a linear graph against t.

Scoring Note: a scale the student’s own values overrun makes correct plotting impossible, so C3 cannot be earned on it.

▸ C4 — 1 point: draws a single straight best-fit line lying among the plotted points, with points falling on both sides of it, and not constrained to pass through the origin or through any single plotted point. Awarded consistent with the points the student actually plotted.

⚠︎ Common error (partial credit): all five points plotted correctly on a workable scale, but joined dot-to-dot as a chain of straight segments rather than fitted with one straight line — earns C3 only, 1 of 2 points.

⚠︎ Common error (partial credit): all five points plotted correctly, but the line is forced through the origin because the ball has fallen no distance at t = 0. The vertical axis is h/t, not h, and its intercept is the launch speed, which is not zero here — earns C3 only, 1 of 2 points.

✗ Common error (no credit): marks a scale that the largest calculated values overrun, so the top points cannot be placed at all, then joins the rest with a smooth curve. With points missing, the data cannot be fitted, and a curve is not a best-fit line.

Part D — Model Answer

Take two widely separated points on the drawn line, not two data points. The line crosses gridline intersections at t = 0.40 s and t = 2.00 s:

slope = (16.0 − 8.0) / (2.00 − 0.40) = 8.0 / 1.60 = 5.0 m/s²

From h/t = u + (g/2)·t the slope is g/2, so

g = 2 × 5.0 = 10 m/s²

Extending the same line back to t = 0 gives a vertical intercept of about 6 m/s, which is the launcher’s speed u. The slope carries the acceleration and the intercept carries the launch speed. The two are independent, and between them they fix the fall time.

That is what settles the students’ argument. Using the measured g = 10 m/s², Ball A takes t_A = 2√(H/g) = 2.83 s to fall 2H = 40 m, while Ball B takes t_B = [−v_0 + √(v_0² + 2gH)]/g = 1.24 s. Ball A does reach the riverbed at the greater speed — 28.3 m/s against Ball B’s 22.4 m/s — but Ball B still arrives first, and it would do so for any throw at all: even at v_0 = 0 a drop from H takes √(2H/g) = 2.0 s, still less than 2.83 s. The speed a ball is moving when it lands is a consequence of its fall, not a cause of an earlier arrival.

Scoring (2 points):

▸ D1 — 1 point: determines the slope of the drawn best-fit line using two points taken on the line rather than two data points. Where the student’s plot matches the model scale this is a value between 4.8 and 5.2 m/s²; otherwise any value consistent with the line the student actually drew, on the scale marked in part C (ii), earns the point, provided the quantity plotted is one whose slope or intercept yields g.

▸ D2 — 1 point: relates the slope to g correctly for the quantity plotted — the slope is g/2 for h/t against t, so g is twice the slope — and evaluates it with the correct unit, m/s². Awarded consistent with the student’s own slope from D1 and with the quantity named in part C (i), whether or not those points were earned.

Scoring Note: a correct, isolated final value of g with the slope marked on the graph earns points D1 and D2. Neither point depends on how close the result lies to any accepted value of g, and the remarks above about the intercept and about the two balls are not required for either.

⚠︎ Common error (partial credit): the slope is determined correctly from the line, but the roles of slope and intercept are exchanged — the intercept is reported as g and the slope as the launcher’s speed — earns D1 only, 1 of 2 points.

✗ Common error (no credit): takes h/t for a single trial, for example 40.0/2.288 = 17.5, and doubles it to give g ≈ 35 m/s². One data point gives the value of the plotted quantity, not the slope of the line; a slope needs the change in that quantity across an interval of t, so no slope has been determined and nothing has been related to one.

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