Attempt each question before opening the hints. Show equations, substitutions, units and the reason for your method. The mark checklists below are PhysicsUK's own schemes.
Lessons from the examiner reports
These reports identify skills worth practising. The questions below are original PhysicsUK examples, with our own mark schemes.
Use the vertical initial velocity in the vertical equation: The report identifies a misconception in which candidates used a horizontal velocity as the initial velocity in a vertical fall calculation. For a horizontal launch, the initial vertical velocity is zero. Both components share the same time but obey their own equations. OCR H156/01, June 2022 — Q23(d); report p. 31
Recombine velocity components at impact: The report emphasises independent horizontal and vertical motion. It describes errors where candidates calculated impact kinetic energy from horizontal speed alone, and where they used distances or energies to find the impact angle. Use the resultant velocity for speed and its components for direction. OCR H156/02, June 2022 — Q4(b)(i)–(iii); report pp. 17–18
Finish a landing calculation with an explicit comparison: The report notes that some candidates found where a ball landed without clearly comparing that position with the full target interval. Show the allowed interval or a distance-from-centre comparison, then give a conclusion supported by your values. OCR H156/01, June 2023 — Q24(b); report p. 36
Explain what the apparatus measures and what the graph means: In the free-fall practical, strong answers specified the height measurement and electronic timing, used points on the fitted line, and related the actual graph axes to the quantity sought. The report also discusses a worst acceptable line for gradient uncertainty. Apply those measurement and analysis habits to the different practical below. OCR H156/02, June 2024 — Q3; report pp. 12–13
Question 1 · Exam-style components and explanation · 5 marks
Velocity at launch and at the highest point
Specification: 2.3.1(d), 3.1.3(a), 3.1.3(b)
A small practice ball is launched at 18.0 m s−1, 40.0° above the horizontal. Take right and up as positive. Neglect air resistance and take g = 9.81 m s−2.
Calculate the initial horizontal and vertical velocity components. [2]
State the speed and the acceleration of the ball at its highest point. Include the direction of the acceleration. [2]
Explain why the horizontal velocity remains constant during the flight. [1]
Small hint
At the highest point only the vertical velocity is zero. The ball is still moving horizontally.
Method hint
The angle is measured from the horizontal: ux = u cos θ and uy = u sin θ. With no air resistance, gravity provides only a vertical force, so ax = 0 and ay = −g.
Complete worked solution and marks
Answer: ux = +13.8 m/s; uy = +11.6 m/s. At the highest point the speed is 13.8 m/s and acceleration is 9.81 m/s² downwards.
ux = 18.0 cos 40.0° = 13.7888… ≈ +13.8 m s−1.Cosine gives the component adjacent to the stated angle.
uy = 18.0 sin 40.0° = 11.5702… ≈ +11.6 m s−1.Use degree mode on the calculator and retain extra digits for later steps.
At the highest point vy = 0, while vx = ux. The speed is therefore √(vx² + 0²) = 13.8 m s−1.
Gravity still acts at the highest point. The acceleration remains 9.81 m s−2 downwards, or ay = −9.81 m s−2 with this sign convention.
There is no horizontal resultant force in the stated model, so the horizontal acceleration is zero and the horizontal velocity remains constant.
Check: √(13.7888…² + 11.5702…²) = 18.0 m/s. Each component is smaller than the launch speed; the apex speed is non-zero.
Mark checklist · 5 marks
1 mark — Horizontal component 18.0 cos 40.0° = +13.8 m/s.
1 mark — Vertical component 18.0 sin 40.0° = +11.6 m/s.
1 mark — Highest-point speed 13.8 m/s, equal to the horizontal component.
1 mark — Highest-point acceleration 9.81 m/s² downwards; accept signed vertical acceleration −9.81 m/s² with up positive.
1 mark — Links zero horizontal resultant force to zero horizontal acceleration and therefore constant horizontal velocity.
Common mistake: Giving zero speed or zero acceleration at the highest point. Vertical velocity reaches zero momentarily, but horizontal velocity and gravity remain.
Exam technique: The 2022 reports support keeping the components separate. PhysicsUK guidance: label the direction of the stated angle before choosing sine or cosine, and distinguish speed, vertical velocity and acceleration at the apex.
The same 18.0 m s−1 launch is now at 50.0° above the horizontal. What happens to the two launch components and the apex speed?
Check the transfer answer
ux = 18.0 cos 50.0° = 11.6 m s−1; uy = 18.0 sin 50.0° = 13.8 m s−1. The components exchange values. The apex speed falls to 11.6 m s−1; its acceleration is still g downwards.
Question 2 · Exam-style horizontal launch · 6 marks
A horizontal launch from a ledge
Specification: 3.1.2(a)(i), 3.1.3(a), 3.1.3(b)
A small ball leaves a horizontal ledge with velocity 4.50 m s−1 to the right. Its centre falls through 1.80 m before its first contact with the level floor. Neglect air resistance, treat the ball as a particle and take g = 9.81 m s−2.
Calculate the time between leaving the ledge and contacting the floor. [2]
Calculate the horizontal distance travelled in that time. [2]
A second ball leaves horizontally at twice the speed from the same position. Explain how its flight time and horizontal distance compare. [2]
Small hint
The initial vertical velocity is zero even though the initial horizontal velocity is 4.50 m/s.
Method hint
Use the vertical drop to find time: 1.80 = ½gt² with down positive. Use that same time in x = uxt. Changing only ux leaves the vertical motion unchanged.
Complete worked solution and marks
Answer: Flight time = 0.606 s; horizontal distance = 2.73 m. Doubling horizontal speed leaves the flight time unchanged and doubles the distance to 5.45 m.
Choose down as positive for this fall: sy = +1.80 m, uy = 0 and ay = +9.81 m s−2.
1.80 = ½ × 9.81 × t² ⇒ t = √(2 × 1.80/9.81) = 0.605782… ≈ 0.606 s.Select the positive time after launch.
x = uxt = 4.50 × 0.605782… = 2.72602… ≈ 2.73 m.The horizontal acceleration is zero, so no horizontal SUVAT acceleration term is needed.
The faster ball has the same vertical initial velocity, vertical drop and gravitational acceleration. It therefore has the same flight time. Its horizontal distance is 9.00 × 0.605782… = 5.45 m, twice the original distance.
Check: After about 0.6 s a free fall covers about ½ × 9.81 × 0.6² = 1.77 m, consistent with the 1.80 m drop. Flight time does not contain ux.
Mark checklist · 6 marks
1 mark — Uses uy = 0 and a valid vertical equation, 1.80 = ½ × 9.81 × t², or the consistent up-positive version.
1 mark — Positive flight time 0.606 s; accept 0.6058 s.
1 mark — Uses x = 4.50t with the vertical flight time; allow error carried forward from a positive part (a) time.
1 mark — Horizontal distance 2.73 m; allow consistent error carried forward.
1 mark — Same flight time, explained by unchanged vertical motion/drop and acceleration.
1 mark — Twice the horizontal distance, explained by doubled horizontal velocity acting for the same time.
Common mistake: Using 4.50 m/s as uy in the vertical equation. That velocity belongs to the horizontal direction; a horizontal launch starts with uy = 0.
Exam technique: This directly practises the independent-components lesson in the 2022 H156/01 report. PhysicsUK guidance: write a separate variable list for each axis, joined by the same time t.
Keep the 4.50 m s−1 horizontal speed but increase the vertical drop to 7.20 m. Find the new flight time and horizontal distance.
Check the transfer answer
The drop is four times larger, so t increases by √4 = 2: 1.21 s. The horizontal distance also doubles to 5.45 m. With ux fixed, both t and x are proportional to √h, not h.
Question 3 · Exam-style equal-height angled launch · 6 marks
Flight time, range and maximum height
Specification: 2.3.1(d), 3.1.3(b)
A ball is launched from level ground at 20.0 m s−1, 35.0° above the horizontal. It first returns to the same level further along the ground. Neglect air resistance, treat the ball as a particle and take g = 9.81 m s−2.
Calculate both initial velocity components. [2]
Determine the non-zero flight time. Show why the equal-height condition matters. [2]
Calculate the horizontal range and maximum height above the launch level. [2]
Small hint
Returning to the same height means sy = 0, not that the ball has travelled no distance vertically.
Method hint
With up positive, solve 0 = uyt − ½gt². The t = 0 solution is launch; the later solution is 2uy/g. Then use R = uxT and H = uy²/(2g).
Complete worked solution and marks
Answer: ux = 16.4 m/s; uy = 11.5 m/s; flight time = 2.34 s; range = 38.3 m; maximum height = 6.71 m.
ux = 20.0 cos 35.0° = 16.3830… m s−1; uy = 20.0 sin 35.0° = 11.4715… m s−1.Rounded answers are 16.4 and 11.5 m s−1; keep the unrounded values in calculations.
Equal launch and landing heights give sy = 0. Thus 0 = uyt − ½gt² = t(uy − ½gt). The two roots are t = 0 at launch and T = 2uy/g at landing.
T = 2 × (20.0 sin 35.0°)/9.81 = 2.338741… ≈ 2.34 s.The ascent and descent times are equal in this equal-height, constant-g model.
R = uxT = (20.0 cos 35.0°) × 2.338741… = 38.3157… ≈ 38.3 m.
At the apex vy = 0. From 0 = uy² − 2gH, H = (20.0 sin 35.0°)²/(2 × 9.81) = 6.70723… ≈ 6.71 m.
Check: Time to the apex is 1.17 s, half the total flight time. The horizontal position of the apex is half the range, about 19.2 m. The equal-height impact speed returns to 20.0 m/s in this ideal model.
Mark checklist · 6 marks
1 mark — ux = 20.0 cos 35.0° = 16.4 m/s.
1 mark — uy = 20.0 sin 35.0° = 11.5 m/s.
1 mark — Uses sy = 0 to obtain 0 = uyt − ½gt² or T = 2uy/g; makes the equal-height condition explicit.
1 mark — Selects the non-zero flight time 2.34 s; allow error carried forward from a resolved positive uy.
1 mark — Range 38.3 m from uxT; allow consistent component/time error carried forward.
1 mark — Maximum height 6.71 m using the vertical component; allow consistent component error carried forward.
Common mistake: Substituting the full 20.0 m/s into a vertical equation, or using T = 2uy/g when launch and landing heights differ. The latter shortcut came from sy = 0.
Exam technique: PhysicsUK guidance: derive the equal-height shortcut from the vertical equation before using it. R = u² sin 2θ/g is a consequence of the same assumptions, not an extra equation that works for every launch.
Try a changed context
Keep the 20.0 m s−1 speed and equal launch/landing heights, but use 55.0°. Compare range, flight time and maximum height.
Check the transfer answer
The range remains 38.3 m because sin 110° = sin 70°. The larger vertical component gives a longer flight time, 3.34 s, and a larger maximum height, 13.7 m. Equal ranges do not imply identical paths.
Question 4 · Exam-style raised launch and quadratic roots · 7 marks
A launch above the landing level
Specification: 2.3.1(d), 3.1.2(a)(i), 3.1.3(b)
A ball is launched at 16.0 m s−1, 30.0° above the horizontal. Its launch point is 4.00 m above the level ground where it lands. Treat it as a particle, neglect air resistance and use g = 9.81 m s−2.
Resolve the launch velocity into horizontal and vertical components. [1]
Form the vertical displacement equation, find both time roots and identify the physically relevant flight time. [4]
Calculate the horizontal range. [1]
Explain why doubling the time to the highest point would give the wrong flight time. [1]
Positions in the diagram use y = 0 at ground level. Vertical displacement from launch to ground is sy = −4.00 m. Open the larger diagramSmall hint
With up positive, the ground is a vertical displacement of −4.00 m from launch. It is not sy = 0.
Method hint
Use ux = 16 cos 30° and uy = 16 sin 30° = 8.00. Solve −4.00 = 8.00t − 4.905t². Keep both roots until you explain which one lies after launch.
Complete worked solution and marks
Answer: ux = 13.9 m/s; uy = 8.00 m/s; roots = −0.401 s and +2.03 s; physical flight time = 2.03 s; horizontal range = 28.2 m.
ux = 16.0 cos 30.0° = 13.8564… m s−1; uy = 16.0 sin 30.0° = 8.00 m s−1.
With up positive, sy = −4.00 m and ay = −9.81 m s−2. Hence −4.00 = 8.00t − 4.905t² ⇒ 4.905t² − 8.00t − 4.00 = 0.
t = [8.00 ± √(8.00² + 4 × 4.905 × 4.00)]/(2 × 4.905).This gives t = −0.401274… s and +2.032262… s.
The model describes the ball after launch at t = 0, so retain the positive root: T = 2.03 s. The negative root belongs to the mathematical extension of the trajectory before launch, not another future impact.
R = uxT = (16.0 cos 30.0°) × 2.032262… = 28.15986… ≈ 28.2 m.
The time to the apex is uy/g = 0.815 s. Doubling it gives 1.63 s, the time to return to the launch height. The actual ground is another 4.00 m lower, so descent takes longer than ascent.
Check: Substituting the unrounded T gives 8T − 4.905T² = −4.00 m. An independent check is ascent time plus fall time from the peak: 8/g + √[2(4.00 + 8²/(2g))/g] = 2.032262… s.
Mark checklist · 7 marks
1 mark — Both resolved components: ux ≈ 13.9 m/s and uy = 8.00 m/s.
1 mark — Correct vertical signs and equation −4.00 = 8.00t − 4.905t², or an equivalent consistently down-positive equation.
1 mark — A valid quadratic method applied to the vertical equation; allow component error carried forward.
1 mark — Both time roots −0.401 s and +2.03 s; allow roots consistent with earlier method error.
1 mark — Selects the positive time and explains that the negative root lies before launch/outside the specified motion.
1 mark — Range 28.2 m from ux times the selected positive flight time; allow consistent error carried forward.
1 mark — Explains that 2uy/g applies to a return to launch height, whereas the lower ground requires a longer descent.
Common mistake: Putting +4.00 into an up-positive displacement equation, discarding a minus sign inside the quadratic, or treating the negative root as a second future landing.
Exam technique: PhysicsUK guidance: make the launch level and landing level visible in your sketch, record the signed vertical displacement and retain unrounded components when solving the quadratic. This raised-launch calculation is an original extension of the independent-components skills in the cited reports.
Keep the same launch components, but move the landing surface to the launch height. Find the new flight time and range.
Check the transfer answer
Now sy = 0. The non-zero time is 2 × 8.00/9.81 = 1.63 s. Range = (16.0 cos 30.0°) × (16.0/9.81) = 22.6 m. This is shorter than the range to the lower ground.
Question 5 · Exam-style two positive roots and landing judgement · 6 marks
Will the ball land on a raised platform?
Specification: 3.1.3(a), 3.1.3(b)
At launch, a small ball has horizontal velocity 10.0 m s−1 rightwards and vertical velocity 9.00 m s−1 upwards. Let x = 0 and y = 0 at launch. A solid platform occupies 13.0 ≤ x ≤ 15.0 m and its horizontal top is 3.00 m above the launch level. Neglect air resistance, treat the ball as a particle and take g = 9.81 m s−2.
Find the two positive times when the ball is at y = 3.00 m. [3]
Find the horizontal position at each of those times. [1]
Decide whether it lands on the platform. Check that it clears the near vertical face at x = 13.0 m as well as meeting the top on its descent. [2]
Dashed line: the height y = 3.00 m. The airborne path ends when the ball first contacts the platform top; the dashed curve beyond that point is the ideal continuation without the platform. Open the larger diagramSmall hint
Here both roots are after launch: the ball passes the same height on the way up and on the way down. Position and direction decide which crossing could be a landing.
Method hint
Solve 3.00 = 9.00t − 4.905t², then x = 10.0t. At the near face t = 13.0/10.0 = 1.30 s; calculate its height there. For the later height crossing, check the interval and the sign of vy = 9.00 − 9.81t.
Complete worked solution and marks
Answer: Height-crossing times = 0.438 s and 1.40 s; positions = 4.38 m and 14.0 m. It clears the near face and lands on the platform at x = 14.0 m while descending.
The vertical equation is 3.00 = 9.00t − 4.905t² ⇒ 4.905t² − 9.00t + 3.00 = 0.
t = [9.00 ± √(9.00² − 4 × 4.905 × 3.00)]/9.81.The roots are 0.437786… s and 1.397076… s. They are both positive and therefore both occur after launch.
At the first crossing x = 10.0 × 0.437786… = 4.38 m, short of the platform. At the second crossing x = 10.0 × 1.397076… = 14.0 m (13.97076… m before rounding).
At the near face x = 13.0 m, t = 1.30 s and y = 9.00 × 1.30 − 4.905 × 1.30² = 3.41 m. This is above the 3.00 m top, so the ball clears the vertical face.
At the later crossing, vy = 9.00 − 9.81 × 1.397076… = −4.71 m s−1. The ball is descending, and 13.0 ≤ 13.97076… ≤ 15.0 m. It therefore lands on the top. The positive early root cannot be discarded just because it is smaller; it describes a real ascent crossing outside the platform.
Check: The apex time is 9.00/9.81 = 0.917 s, between the two roots. Its maximum rise is 9.00²/(2 × 9.81) = 4.13 m, so two crossings of 3.00 m are possible. The height at the near face exceeds the top by about 0.41 m.
Mark checklist · 6 marks
1 mark — Correct signed height equation 3.00 = 9.00t − 4.905t².
1 mark — Correct quadratic method, retaining both solutions.
1 mark — Two positive roots 0.438 s and 1.40 s; accept unrounded equivalents.
1 mark — Both horizontal positions 4.38 m and 14.0 m from x = 10.0t; allow time error carried forward.
1 mark — Checks the near face: y(1.30 s) = 3.41 m > 3.00 m, so it clears the face.
1 mark — Checks the later position lies within 13.0–15.0 m and the ball is descending, then concludes it lands on the top. Accept a negative vy, or a correct comparison with the apex time, as evidence of descent.
Common mistake: Discarding every smaller root, or checking only the final x value for a solid platform. A raised level can be crossed twice after launch; a platform face can intercept a trajectory before it reaches the calculated top crossing.
Exam technique: The 2023 report supports explicitly comparing a landing position with the full target interval. PhysicsUK extension: for a raised solid obstacle, also check its near face and whether the relevant crossing is on descent.
The platform is moved to 14.5 ≤ x ≤ 16.5 m but retains its 3.00 m top height. Does the ball land on its top?
Check the transfer answer
No. The descent crossing at x = 13.97076… m is before the platform starts. At its near face, t = 1.45 s and y = 9.00 × 1.45 − 4.905 × 1.45² = 2.74 m, below the top. It strikes the vertical face, rather than landing on the top.
Question 6 · Exam-style impact velocity and energy check · 6 marks
A 0.150 kg ball is fired horizontally at 8.00 m s−1. Its centre falls through 5.00 m before impact. Neglect air resistance, treat the ball as a particle and take g = 9.81 m s−2.
Calculate its vertical velocity immediately before impact, taking up as positive. [2]
Calculate the impact speed. [2]
Calculate the direction of the impact velocity as an angle below the horizontal. [1]
Write an energy calculation that checks the impact speed. [1]
Diagram not to scale. Use a velocity triangle for the impact angle. The slope of the line joining launch and landing positions gives an average path angle, which is a different quantity. Open the larger diagramSmall hint
The horizontal velocity is still 8.00 m/s at impact, but the vertical velocity is no longer zero.
Method hint
Find vy² = 2g × 5.00 and choose the downward sign. Use v = √(vx² + vy²) and β = tan⁻¹(|vy|/vx). The independent energy check is ½mv² = ½m(8.00)² + mg(5.00).
Complete worked solution and marks
Answer: vy = −9.90 m/s; impact speed = 12.7 m/s; direction = 51.1° below horizontal. Initial kinetic energy 4.80 J plus 7.36 J gravitational energy gives 12.2 J at impact.
With up positive, uy = 0, sy = −5.00 m and ay = −9.81 m s−2. vy² = 0 + 2(−9.81)(−5.00) = 98.1 m² s−2.
The ball moves downwards immediately before impact, so vy = −√98.1 = −9.90454… ≈ −9.90 m s−1. The positive square root alone gives a magnitude, not the signed component.
v = √[8.00² + (−9.90454…)²] = √162.1 = 12.73184… ≈ 12.7 m s−1.Do not add perpendicular components arithmetically.
The initial kinetic energy is ½ × 0.150 × 8.00² = 4.80 J. The gravitational energy transferred is 0.150 × 9.81 × 5.00 = 7.3575 J. Thus ½ × 0.150 × v² = 4.80 + 7.3575 = 12.1575 J, giving v = 12.7 m s−1 independently. The mass cancels when solving for speed.
Check: The impact speed exceeds each component and the 8.00 m/s launch speed. The angle is larger than 45° because the vertical speed exceeds the horizontal speed. Kinetic energy increases by mgh, rather than becoming mgh alone.
1 mark — Impact angle 51.1° below horizontal using velocity components; allow consistent error carried forward.
1 mark — Energy check includes both initial horizontal kinetic energy and gravitational energy: ½ × 0.150 × v² = ½ × 0.150 × 8.00² + 0.150 × 9.81 × 5.00, obtaining the same speed. Accept the mass-cancelled equation v² = 8.00² + 2g × 5.00.
Common mistake: Using only horizontal speed for impact energy, using only mgh as the final energy, or finding the impact angle from drop/range. The trajectory is curved, so the displacement ratio does not give its tangent direction at impact.
Exam technique: This practises both reported errors in 2022 H156/02 Q4(b): omitting the gained vertical contribution to kinetic energy and using positions or energies instead of velocity components for direction.
Double the ball's mass while keeping its launch velocity and vertical drop unchanged. What happens to flight time, impact speed and impact kinetic energy in this model?
Check the transfer answer
Flight time and impact speed are unchanged because the kinematic equations contain g but not mass. Impact kinetic energy doubles to 24.3 J. This comparison assumes air resistance remains negligible.
Question 7 · Exam-style reverse design and comparison · 5 marks
Choose a launch speed to reach a target
Specification: 3.1.3(a), 3.1.3(b)
A horizontal launcher must send a particle-sized ball to a point 3.00 m horizontally from its launch position. The vertical drop from launch to impact is 1.25 m. Neglect air resistance and take g = 9.81 m s−2.
Calculate the required horizontal launch speed, showing how you obtain the flight time. [3]
The launcher is then raised so that the drop is 2.50 m, while its horizontal speed remains unchanged. Calculate the new range and explain the scale factor. [2]
Small hint
The target gives distance, not flight time. Find time from the vertical drop before working backwards to speed.
Method hint
For a horizontal launch t = √(2h/g), then ux = R/t. If h doubles while ux stays fixed, t and R increase by √2.
Complete worked solution and marks
Answer: Required horizontal speed = 5.94 m/s. Doubling the drop gives a new range of 4.24 m, √2 times the original range.
t = √(2h/g) = √(2 × 1.25/9.81) = 0.504818… s.The initial vertical velocity is zero.
ux = R/t = 3.00/0.504818… = 5.94272… ≈ 5.94 m s−1.This is the required speed for the ideal model.
At the doubled drop, tnew = √(2 × 2.50/9.81) = √2 × t.Since ux is fixed, Rnew = uxtnew = √2 × 3.00 = 4.24 m.
To keep the original 3.00 m target distance at the greater drop, the launcher would instead need ux,new = 5.94272…/√2 = 4.20 m s−1. More flight time means a smaller required horizontal speed for a fixed target.
Check: R/ux has units of seconds and agrees with the vertical flight time. Doubling height does not double time because h is proportional to t².
Mark checklist · 5 marks
1 mark — Uses t = √(2h/g) with uy = 0 and h = 1.25 m.
1 mark — Uses ux = 3.00/t with a positive flight time; allow time error carried forward.
1 mark — Explains the √2 increase in flight time and therefore range when h doubles and ux is fixed.
1 mark — New range 4.24 m; accept 3.00√2 m or consistent unrounded calculation.
Common mistake: Doubling range just because the drop doubles, or using R = u²/g for a horizontal raised launch. That range shortcut is for a 45° equal-height launch, not this geometry.
Exam technique: PhysicsUK guidance: state the design target, calculate the common flight time from the vertical motion, then rearrange the horizontal equation. Real launchers require calibration; the ideal prediction is a starting value.
At the original 1.25 m drop, the acceptable landing interval is 2.90–3.10 m. What horizontal speed interval does the ideal model allow?
Check the transfer answer
Use the same t = 0.504818… s for both bounds. Speeds run from 2.90/t = 5.74 m s−1 to 3.10/t = 6.14 m s−1. A proposed speed should be compared with both limits.
Question 8 · Exam-style practical graph and model evaluation · 8 marks
A student launches a small ball horizontally at several speeds u. The vertical drop of the ball's centre to first contact is fixed at h = 0.800 m. The student records horizontal range R and plots R vertically against u horizontally. For the ideal prediction use g = 9.81 m s−2 and neglect air resistance.
Mean results from repeated launches
u / m s−1
R / m
2.00
0.814
3.00
1.194
4.00
1.594
5.00
2.014
Two exact construction points on the best-fit line are A = (2.00 m s−1, 0.804 m) and B = (5.00 m s−1, 2.004 m). These are fitted-line points, not individual readings. A worst acceptable line gives gradient 0.410 s; assume the difference from the best gradient represents the absolute gradient uncertainty.
Derive the ideal relationship between R and u and calculate its predicted gradient. [2]
Use A and B to calculate the experimental gradient with its unit. [2]
Use the stated uncertainty rule to decide whether the experimental gradient is consistent with the ideal prediction. [1]
Describe how to measure launch speed, the centre's vertical drop and the mean horizontal range reliably. Give apparatus and measurement details, rather than just naming instruments. [3]
Crosses show the measured means; A and B are construction points on the fitted line. Exact coordinates are supplied in the question. Open the larger diagramSmall hint
The gradient is range divided by speed: metres divided by metres per second. It has the unit of time.
Method hint
Vertical motion gives t = √(2h/g). Substitute into R = ut, so the theoretical gradient is the flight time. Calculate (RB − RA)/(uB − uA) for the fitted gradient. Compare the predicted value with 0.400 ± |0.410 − 0.400| s.
Complete worked solution and marks
Answer: Ideal relation R = u√(2h/g); predicted gradient = 0.404 s. Measured gradient = 0.400 s with uncertainty ±0.010 s, consistent with the ideal prediction within this stated gradient uncertainty.
For a horizontal launch uy = 0. Vertical motion gives h = ½gt² and t = √(2h/g). Horizontal motion gives R = ut = u√(2h/g).At fixed h the ideal graph is a straight line through the origin.
Predicted gradient = √(2 × 0.800/9.81) = 0.403855… ≈ 0.404 s.The coefficient is the vertical flight time, not the horizontal speed or g.
Use the line points, not a single R/u ratio: m = (2.004 − 0.804)/(5.00 − 2.00) = 1.200/3.00 = 0.400 s.The unit is m/(m s−1) = s. The line's small 0.004 m intercept does not change its gradient.
The stated uncertainty rule gives Δm = |0.410 − 0.400| = 0.010 s. The interval is 0.390–0.410 s, containing the ideal 0.403855… s. The result is consistent with the ideal prediction within this stated gradient uncertainty; it does not prove air resistance is exactly zero or rule out other systematic errors.
Launch speed: place a light gate at the horizontal exit and connect it to an electronic timer or data logger. Measure the ball's diameter and divide it by the beam-interruption time, keeping the beam through the centre and the gate close enough to the exit that this is the horizontal launch speed. Alternatively, use calibrated video with a horizontal scale and known frame rate to track the centre just after exit.
Height: clamp the launcher so the exit is horizontal and use a vertical rule to measure the fall of the ball's centre. For a ball of radius r landing on a horizontal surface, the centre at first contact is r above that surface; measure centre-at-launch to centre-at-contact, or use the equivalent exit-support-to-floor height when the geometry matches. Keep this drop fixed as speed is varied.
Range: use a plumb line from the launch centre to mark its vertical projection on paper at the landing surface. Use suitable impact-marking paper over a protected landing surface and measure horizontally to the first impact mark. Repeat launches at each speed, find mean R and record the spread. Keep the launch direction fixed, avoid parallax and use a clear, contained landing area.
Check: The predicted gradient is the same flight time for every launch at this fixed drop. The measured 0.400 s differs by about 1% from 0.403855… s, less than the stated 2.5% gradient uncertainty. The best-fit construction does not force the intercept to zero.
Mark checklist · 8 marks
1 mark — Derives R = u√(2h/g) from uy = 0, h = ½gt² and R = ut.
1 mark — Predicted gradient √(2 × 0.800/9.81) = 0.404 s.
1 mark — Uses both fitted-line points: (2.004 − 0.804)/(5.00 − 2.00).
1 mark — Measured gradient 0.400 s with the time unit; no credit here for 0.400 m/s or m/s².
1 mark — Computes ±0.010 s, checks 0.404 s lies within 0.390–0.410 s and concludes consistency within the stated gradient uncertainty.
1 mark — Usable launch-speed method with apparatus and analysis: a centred light gate at the exit, known ball diameter and diameter divided by electronic interruption time; or calibrated centre-tracking video with known frame rate and displacement/time.
1 mark — Measures and fixes the centre's vertical drop with a vertical rule, accounting for the ball's radius/first-contact position, and maintains a horizontal launch.
1 mark — Marks the vertical projection of launch, measures to first impact horizontally, and repeats/means ranges at each speed. A bare ‘measure range with a ruler’ is insufficient.
Common mistake: Using one measured R/u as the fitted gradient, taking points off the fitted line, assigning speed units to the gradient, or saying ‘repeat’ without identifying which measurement is repeated and how it is analysed.
Exam technique: The 2024 report concerns a different free-fall investigation. Its transferable lessons are to explain the apparatus and measurements, use fitted-line points and connect the actual axes to the model. Our eight-mark point checklist is PhysicsUK's scheme; the source practical used OCR levels of response.
If air resistance is appreciable, why does the horizontal equation R = ut cease to be exact? Does the small difference between the two gradients above prove that drag caused it?
Check the transfer answer
Drag has a horizontal component opposing the motion, so vx decreases and the horizontal distance is the area under the vx–t graph, not initial u multiplied by the entire flight time. Drag can also change the vertical motion. The measured gradient difference is inside the stated uncertainty, so it does not establish a drag effect; height, speed and range errors also need consideration.
Use the simulation to check your predictions
Set Angle to 30°, Speed to 16 m/s, Height to 4 m and Gravity to 9.81. This recreates Question 4. Predict a flight time of 2.03 s and range of 28.2 m. Use Reset, then Step +0.1 s ten times: at t = 1.00 s expect vx = 13.86 m/s and vy = −1.81 m/s. The projectile and the vertical-only ball should have the same height at every time because they have the same initial vertical velocity and acceleration. The vertical-only ball is launched upwards here; it is not dropped from rest. Continue stepping until impact and compare the Flight time and Range metrics with your prediction. Finally set Angle to 0°: both balls now start with zero vertical velocity and reach the ground together. The explorer models ideal motion with no air resistance.
Written for OCR A AS Physics H156 projectile motion, 3.1.3(a)–(b), shared with the first year of H556. Supporting skills include resolving vectors, constant-acceleration equations, graph gradients, practical measurement and uncertainty. The linked A-level notes cover this shared content. Each calculation specifies g and the no-air-resistance model. Range shortcuts are derived for equal launch and landing heights only. The set includes height differences, two positive times at a raised level, impact direction, a reverse-design calculation and a practical model check. Collisions, spin, lift and numerical drag modelling require separate practice; the ideal model does not predict a precise real-world landing position when those effects matter.
The question index was used to locate assessment examples, then the actual questions, mark schemes and examiner reports were checked. The report lessons above are paraphrases. Their application to these original questions is PhysicsUK teaching guidance, not an OCR mark scheme or a prediction of future exam frequency. The practical graph here is a new horizontal-launch investigation; the linked 2024 report discusses a different free-fall practical and supports the transferable measurement and fitted-line skills. Read the linear-motion set first if SUVAT signs or gradients are unfamiliar.
