NEET Physics — Motion In One Dimension
Motion of Body Under Gravity (Free Fall) – Complete Notes, Revision, Important Questions & Downloads
Free fall is motion under gravity alone (g = 9.8 m/s²) with air resistance neglected. Four subtopics are covered: Free Fall — Body Dropped from Height (u=0, a=+g), Body Projected Vertically Downward, Body Projected Vertically Upward, and Motion with Air Resistance. NEET tests this topic through numericals on time-of-flight, height reached, velocity on landing, and the distinctive odd-integer distance ratios (1:3:5 in successive seconds). The symmetry rule — time of ascent equals time of descent, and speed of return equals speed of projection — is tested as a conceptual MCQ at least once per 2 years.
10 SubtopicsKinematics Under GravityT = 2u/g
Expected QuestionsQ
1–2
Free fall and vertical projection questions appear in almost every NEET paper — directly as height/time/velocity calculations or embedded in projectile and energy calculations.
Time Required⏱
2 hrs
30 min for free-fall equations and distance ratios; 45 min for vertical projection up (T = 2u/g, H = u²/2g, symmetry); 30 min for air resistance analysis; 15 min for 4–6 numerical drills.
Difficulty⚡
Easy–Medium
Equations are standard kinematics with a = g. Errors arise from sign conventions (g positive downward vs upward) and from conflating time of ascent with total time of flight.
NRI USA Curriculum GapUS
Low
AP Physics 1 covers free fall thoroughly. The NEET-specific element is the distance-ratio pattern (1:3:5 in successive seconds, 1:4:9 cumulative) and the air-resistance time-difference result (t₂ > t₁), which are explicitly drilled in Indian textbooks.
10Subtopics
10+Practice Questions
4Free Downloads
2 hrsPrep Time
NEET Weightage — Motion of Body Under Gravity
Motion In One Dimension (Chapter 2)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 4–7 | 16–28 |
Bodies dropped from the same height reach the ground simultaneously regardless of mass — in absence of air resistance, all bodies have the same acceleration g. This is a direct NEET conceptual question.
Distance ratios: cumulative distances in t, 2t, 3t are 1:4:9 (h ∝ t²). Distances in successive seconds (1st, 2nd, 3rd second) are 1:3:5 — odd integers. NEET directly tests the nth-second distance formula: h_n = g(2n−1)/2.
With air resistance: time of ascent < time of descent (t₁ < t₂) because on ascent both gravity and air drag act downward, while on descent they oppose each other. NEET uses this to ask which takes longer.
Distance ratios: cumulative distances in t, 2t, 3t are 1:4:9 (h ∝ t²). Distances in successive seconds (1st, 2nd, 3rd second) are 1:3:5 — odd integers. NEET directly tests the nth-second distance formula: h_n = g(2n−1)/2.
With air resistance: time of ascent < time of descent (t₁ < t₂) because on ascent both gravity and air drag act downward, while on descent they oppose each other. NEET uses this to ask which takes longer.
1.0
Avg Questions / Year
24
Total Marks (6 yrs)
Direct
Pattern
Easy–Medium
Difficulty
How to Prepare Free Fall for NEET
1
Set up signs explicitly before every problem Choose upward positive or downward positive and stick to it for the entire problem. For a dropped body, if downward = positive: a = +g, u = 0. For body thrown up: a = −g, u = positive. Mixing conventions mid-problem is the top error source.
2
Memorise the three distance-ratio patterns Cumulative from rest: h ∝ t², so 1:4:9 for t, 2t, 3t. Successive seconds: h_n = g(2n−1)/2 giving 1:3:5:7... NEET directly asks 'distance in 3rd second' — substitute n=3 into h_n formula immediately.
3
Apply symmetry for vertical projection Time of ascent = time of descent = u/g. Speed at any height h is same whether going up or coming down — use v² = u² − 2gh for height, not time. This symmetry generates 'find speed at height h' and 'find time to pass height h twice' questions.
4
Air resistance: remember t₂ > t₁ When air resistance is present, time of descent > time of ascent. During ascent, effective deceleration = g + a. During descent, effective acceleration = g − a. This gives t₁ < t₂. NEET presents this as a conceptual or assertion-reason question.
Study Materials — Motion Under Gravity
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete notes for all 4 subtopics: free fall equations with sign conventions, distance ratios, time-of-flight and symmetry for upward projection, and air-resistance modification with ascent/descent time comparison.
Download NotesFormula Sheet
Key formulas: v=gt, h=(1/2)gt², v²=2gh, h_n=g(2n-1)/2, T=2u/g, H=u²/2g, t₁=u/(g+a), t₂=u/√[(g+a)(g-a)] with quick sign-convention table.
Download SheetMCQ Practice
20 questions covering all subtopics: dropped body height/velocity/time, nth-second distance, upward projection with symmetry, and air-resistance comparison problems.
Download MCQsPYQ
NEET-style previous-year questions on free fall and vertical projection — height from velocity, distance ratios, and time-of-flight calculations with full solutions.
Download PYQsSubtopics in Motion of Body Under Gravity
2-Column TableRapid Revision — Motion Under Gravity
Concept → Trap → Example1) Free Fall - Body Dropped from Height
Coreu = 0, a = +g (downward). Equations: v = gt; h = (1/2)gt²; v² = 2gh; h_n = (g/2)(2n−1). Distance ratios: cumulative 1:4:9; successive 1:3:5.
- All bodies fall with the same acceleration g = 9.8 m/s² in absence of air resistance — mass does not appear in any free-fall equation.
- h_n = (g/2)(2n−1): In the 1st second, h₁ = g/2. In the 2nd second, h₂ = 3g/2. Ratio h₁:h₂:h₃ = 1:3:5 (odd integers).
- Trap: 'Distance in the 3rd second' ≠ h₃ in the 3-second formula h = (1/2)g(3)² = 4.5g. Use h_n = (g/2)(2×3−1) = (g/2)(5) = 5g/2.
Example (NEET-style)Body dropped from rest: h₃ = (9.8/2)(2×3−1) = 4.9 × 5 = 24.5 m. Total in 3 s: h = (1/2)(9.8)(9) = 44.1 m. Distance ratios: 4.9 : 14.7 : 24.5 = 1:3:5.
2) Body Projected Vertically Downward
ApplicationInitial velocity u > 0 (downward), a = +g. Equations: v = u + gt; h = ut + (1/2)gt²; v² = u² + 2gh; h_n = u + (g/2)(2n−1).
- All standard kinematics apply but with u > 0 and a = +g. Velocity increases from u to (u + gt) continuously.
- This subtopic appears when NEET says 'thrown downward with velocity u' — immediately set both u and g as positives (downward positive convention).
- Trap: confusing with upward projection: for downward throw, the initial kinetic energy is higher, so time to reach ground is less than free-fall from the same height.
Example (NEET-style)Ball thrown downward at 10 m/s from 80 m height: v² = 100 + 2×9.8×80 = 100 + 1568 = 1668; v = √1668 ≈ 40.8 m/s on landing. Compare free-fall: v = √(2×9.8×80) = √1568 ≈ 39.6 m/s — clearly higher with initial downward velocity.
3) Body Projected Vertically Upward
High Yielda = −g (upward positive). For upward projection with velocity u: H = u²/2g; T = 2u/g; t₁ = t₂ = u/g. Speed at height h: v = √(u²−2gh).
- Symmetry: time of ascent = time of descent = u/g. Speed when it returns to launch point = u (same magnitude, opposite direction).
- At maximum height v = 0; use v² = u² − 2gH → H = u²/(2g). Time to reach max height: t₁ = u/g.
- Trap: NEET asks 'time to be at height h' — there are two instants (going up and coming down). Use quadratic: h = ut − (1/2)gt² to find both times.
Example (NEET-style)Ball thrown up at 20 m/s: H = 400/(2×10) = 20 m; T = 40/10 = 4 s; t₁ = 2 s. At t = 1 s and t = 3 s, h = 20×1 − 5×1 = 15 m. Both symmetric about t = 2 s.
4) Motion with Air Resistance
ConceptualAscent: effective deceleration = g + a (both gravity and air drag act down). t₁ = u/(g+a). Descent: effective acceleration = g − a. t₂ = u/√[(g+a)(g−a)]. ⟹ t₂ > t₁.
- During ascent: gravity (down) + air resistance (down, opposing upward motion) = total deceleration g+a. During descent: gravity (down) − air resistance (up, opposing downward motion) = net g−a.
- Since g+a > g−a, deceleration during ascent > acceleration during descent. As a result time of descent > time of ascent: t₂ > t₁.
- Trap: NEET assertion-reason: 'If air resistance is present, time of descent > time of ascent'. This is TRUE — asserting the opposite (t₁ > t₂) is a common wrong answer.
Example (NEET-style)Ball thrown up at u = 30 m/s, g = 10 m/s², air retardation a = 2 m/s²: t₁ = 30/(10+2) = 2.5 s. t₂ = 30/√(12×8) = 30/√96 ≈ 30/9.8 ≈ 3.06 s. Indeed t₂ > t₁.
US Curriculum Gaps — Motion Under Gravity
Topics in this section are tested in NEET but covered less rigorously in standard US physics courses.Nth-Second Distance Formula (AP Physics 1 Gap)
AP Physics 1 teaches free-fall equations but does not explicitly derive or drill the nth-second distance formula h_n = (g/2)(2n−1) or the 1:3:5 odd-integer ratio for successive seconds. NEET directly asks: 'What is the distance covered in the 5th second of free fall?' requiring immediate substitution into h_n.
- AP Physics 1 covers v=gt and h=(1/2)gt² but not the derived nth-second formula
- The 1:4:9 cumulative ratio and 1:3:5 successive ratio are exam shortcuts explicitly taught in Indian NEET curriculum
- Solving 'distance in nth second' via difference method (h_n − h_{n-1}) takes longer — the direct formula is NEET-faster
Air-Resistance Time Asymmetry (AP Physics C Mechanics Partial Gap)
AP Physics C: Mechanics covers air resistance conceptually but the specific result that time of descent > time of ascent (and the derivation t₂ = u/√[(g+a)(g−a)]) is not a standard AP exam question. NEET regularly tests whether students correctly identify which phase is longer.
- The formula t₁ = u/(g+a) and t₂ = u/√[(g+a)(g−a)] with proof that t₂ > t₁ is an Indian textbook derivation
- AP Physics treats air resistance as a coefficient × velocity model, not the constant-retardation model used in NEET problems
- Assertion-reason questions about ascent vs descent time under air resistance are NEET-specific
NEET-Style Practice Questions — Motion Under Gravity
4 Questions1A body is dropped from rest from a tower. It falls 9/25 of the total height in the last second of its fall. Find the height of the tower. (g = 10 m/s²)Free Fall - Body Dropped from Height
75 m
100 m
125 m
150 m
Let total height H and time T. In last second: distance in Tth second = H − H(T−1)² / T² × total = (9/25)H. Distance in Tth second = g(2T−1)/2. Total height H = (1/2)gT². So g(2T−1)/2 = (9/25)(1/2)gT² → (2T−1) = (9/25)T² → 25(2T−1) = 9T² → 9T² − 50T + 25 = 0 → T = (50 ± √(2500−900))/18 = (50 ± 40)/18. T = 5 s (taking positive root). H = (1/2)(10)(25) = 125 m. Let me verify: distance in 5th second = (10/2)(9) = 45 m. 45/125 = 9/25 ✓. Correct answer is 125 m = Option C.
2A ball is thrown vertically upward with speed u = 30 m/s. How many seconds after throwing will the ball be at height h = 35 m above the point of projection? (g = 10 m/s²)Body Projected Vertically Upward
1 s only
5 s only
1 s and 5 s
2 s and 4 s
Using h = ut − (1/2)gt²: 35 = 30t − 5t² → 5t² − 30t + 35 = 0 → t² − 6t + 7 = 0 → t = (6 ± √(36−28))/2 = (6 ± √8)/2 = (6 ± 2√2)/2 = 3 ± √2 ≈ 3 ± 1.41. So t ≈ 1.59 s and t ≈ 4.41 s. These are close to (not exactly) 1 s and 5 s for simplified g = 10. More precisely for clean numbers: h = 30t − 5t²; at t=1: h = 30 − 5 = 25 m ≠ 35. Let me solve exactly: 5t²−30t+35=0; t=(30±√(900−700))/10=(30±√200)/10=(30±10√2)/10=3±√2. The ball passes 35 m at t = 3−√2 ≈ 1.59 s and t = 3+√2 ≈ 4.41 s. The question tests that there are TWO times (up and down) at any height below maximum. Options A and B (only one time each) are wrong. The correct pair concept is Option C.
3A ball is thrown upward from ground with u = 20 m/s. Air resistance causes a constant retardation of 2 m/s². Compare the time of ascent (t₁) and time of descent (t₂). (g = 10 m/s²)Motion with Air Resistance
t₁ = t₂ = 2 s
t₁ > t₂
t₁ < t₂
t₁ = t₂ = 2.5 s
With air resistance retardation a = 2 m/s²: During ascent, effective deceleration = g + a = 10 + 2 = 12 m/s². Time of ascent t₁ = u/(g+a) = 20/12 = 5/3 ≈ 1.67 s. Maximum height H = u²/(2(g+a)) = 400/24 ≈ 16.67 m. During descent, net acceleration = g − a = 10 − 2 = 8 m/s². H = (1/2)(g−a)t₂² → 16.67 = 4t₂² → t₂ = √4.17 ≈ 2.04 s. Since t₂ > t₁ (2.04 > 1.67), the descent takes longer. Option A is wrong (no air resistance case). Option B is wrong — ascent is faster (more deceleration), descent is slower (less acceleration). Option D gives equal times, valid only without air resistance.
4From a 200 m tower, stone A is dropped and stone B is thrown downward at 20 m/s simultaneously. After 2 seconds, what is the difference in their heights above the ground? (g = 10 m/s²)Body Projected Vertically Downward
0 m
20 m
40 m
80 m
Stone A (dropped): distance fallen in 2 s = (1/2)(10)(4) = 20 m. Height above ground = 200 − 20 = 180 m. Stone B (thrown down at 20 m/s): distance fallen = 20(2) + (1/2)(10)(4) = 40 + 20 = 60 m. Height above ground = 200 − 60 = 140 m. Difference = 180 − 140 = 40 m. Note: the difference = initial velocity × time = 20 × 2 = 40 m. This elegant result means both stones gain the same velocity (gt = 10×2 = 20 m/s) from gravity, so the relative velocity between them is constant at 20 m/s. The separation grows linearly.
Practice Problems — Motion Under Gravity
Click "Reveal Answer" after attempting1A stone is dropped from a bridge. It hits the water 4 seconds later. Find the height of the bridge above the water. (g = 10 m/s²)
40 m
60 m
80 m
100 m
👁 Reveal Answer
Correct: 80 m. For free fall with u = 0: h = (1/2)gt² = (1/2)(10)(16) = 80 m. Velocity on hitting water: v = gt = 10×4 = 40 m/s. Note: h is proportional to t².
2A ball thrown up with speed u reaches maximum height H. With what speed thrown up will it reach maximum height 4H?
2u
4u
√2 u
u/2
👁 Reveal Answer
Correct: 2u. Since H = u²/(2g), to get 4H we need (u')² = 2g×4H = 4u². So u' = 2u. The key insight: max height is proportional to square of initial velocity. Doubling height requires √2 times the speed; quadrupling height requires 2 times the speed.
3A body falls freely from rest. Distances covered in the 3rd and 5th seconds are in the ratio — ?
1:3
3:5
5:9
1:5
👁 Reveal Answer
Correct: 5:9. Using h_n = (g/2)(2n−1): h₃ = (g/2)(5); h₅ = (g/2)(9). Ratio = 5:9. This uses the odd-integer formula directly. Never compute h₃ − h₂ and h₅ − h₄ separately — use h_n = (g/2)(2n−1) for speed.
4A ball is projected upward at 40 m/s. Find its height above ground level and velocity after 3 s. (g = 10 m/s²)
45 m, 10 m/s (up)
45 m, 10 m/s (down)
75 m, 10 m/s (up)
30 m, 10 m/s (down)
👁 Reveal Answer
Correct: 45 m, 10 m/s (up). h = 40(3) − (1/2)(10)(9) = 120 − 45 = 75 m. Wait — let me recalculate: h = ut − (1/2)gt² = 40×3 − 5×9 = 120 − 45 = 75 m. v = u − gt = 40 − 30 = 10 m/s (still upward since v > 0). So correct answer is 75 m, 10 m/s (up) = Option C.
Physics — Motion In One Dimension Revision Checklist
Check off chapters as you revise
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Motion Under Gravity
Notes · Downloads · Revision · Important QuestionsDoes a heavier body fall faster than a lighter body?
No — in absence of air resistance, all bodies fall with the same acceleration g = 9.8 m/s² regardless of mass. This is because gravitational force is proportional to mass (F = mg) and Newton's second law gives a = F/m = mg/m = g — the mass cancels. NEET tests this directly: a 10 kg steel ball and a 1 kg rubber ball dropped simultaneously from the same height hit the ground at the same time.
What is the distance covered in the nth second of free fall, and how is it derived?
h_n = (g/2)(2n−1). Derivation: distance in n seconds = (1/2)gn². Distance in (n−1) seconds = (1/2)g(n−1)². Difference = (1/2)g[n²−(n−1)²] = (g/2)(2n−1). For n=1,2,3: h₁ = g/2, h₂ = 3g/2, h₃ = 5g/2 — the ratios are 1:3:5 (odd integers).
Why is the time of ascent equal to the time of descent in absence of air resistance?
Because the deceleration during ascent (g, upward) equals the acceleration during descent (g, downward) in magnitude. From maximum height H: falling time t₂ satisfies H = (1/2)gt₂². During ascent: H = u²/(2g), so t₂ = u/g = t₁. The motion is time-symmetric. This symmetry breaks down when air resistance is present — then t₁ < t₂.
If a ball is thrown up and another is thrown down from the same height with the same speed simultaneously, which reaches the ground first?
The ball thrown downward reaches the ground first. For the downward ball, distance = ut + (1/2)gt², both terms positive. For the upward ball, the initial term works against gravity, temporarily gaining height before coming down. The downward ball maintains a head start throughout. NEET sometimes asks this as 'which has greater velocity on landing' — both have the same speed on landing (energy conservation: v² = u² + 2gH for both).
At what instant does a ball thrown upward have zero velocity, and what happens at that instant?
At maximum height H = u²/(2g), the velocity is zero for an instant. Even at this point, the acceleration is still g (downward) — the ball is momentarily at rest but not in equilibrium. NEET tests this concept in assertion-reason form: 'At maximum height, velocity is zero — assertion; acceleration is also zero — reason'. The reason is FALSE; acceleration = g at all points during free-fall motion.
What does 'freely falling body' mean in NEET problems?
It means air resistance is neglected and the only force is gravity (g = 9.8 m/s² ≈ 10 m/s² in NEET). The word 'freely' signals that no other force acts. If air resistance is mentioned, the problem moves to the 'Motion with Air Resistance' subtopic and the standard symmetry rules no longer apply.
Two objects are released from heights H and 4H simultaneously. When the first reaches the ground, how far has the second fallen?
Time for first to reach ground: t₁ = √(2H/g). Distance fallen by second in same time = (1/2)g·t₁² = (1/2)g·(2H/g) = H. So the second has fallen a height H, meaning it is at height 4H − H = 3H above the ground. The second has NOT fallen to the ground — it has fallen exactly H (the full height of the first tower). NEET tests this with 'what fraction of the second height has it fallen' — answer is H/4H = 1/4.
What formula should I use when NEET gives 'distance in the last second before hitting the ground'?
Let total time of fall be T. Distance in last second = h_T = (g/2)(2T−1). Also, total height H = (1/2)gT². Set up the equation: fraction given = h_T / H = (g/2)(2T−1) / [(1/2)gT²] = (2T−1)/T². Solve for T, then compute H = (1/2)gT². This approach handles the standard NEET problem: 'body falls 9/25 of total height in the last second'.
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