Motion under gravity: free fall and vertical projection
The kinematic equations with a = −g. The physics is short; the sign convention is everything.
NCERT Ex. 2.3–2.5, 2.7Priority 120 worked questions2 PYQs3 animated figures
1 · The concept in plain language
The one idea
Free fall is not a new set of equations. It is the kinematic equations from File 2 with
a = −g substituted in. Everything difficult about this topic is
sign convention, not physics.
An object released near the Earth’s surface accelerates downward under gravity.
If air resistance is neglected it is said to be in free fall. Provided the height of the fall is
small compared with the Earth’s radius, g can be treated as a constant, 9.8 m s⁻². Free
fall is therefore simply a case of motion with uniform acceleration.
Note what free fall does not require: it does not require the body to start from rest, and it
does not care about the body’s mass. A ball thrown straight up is in free fall from the instant it
leaves your hand — including at the top, where its velocity is zero but its acceleration is still
9.8 m s⁻² downward.
Setting up the sign convention
NCERT takes the y-axis vertically upward as positive. Then gravity, always pointing down, is
a = −g = −9.8 m s⁻². Two working rules:
If everything in the problem moves downward (a dropped stone, a bullet fired down), take
downward positive. Every number is then positive and no sign errors can occur.
If anything goes up, take upward positive and write g as −9.8 (or −10).
Then never change convention mid-problem, no matter how tempting.
animated NCERT Example 2.3, animated on the real equation y = 25 + 20t − 5t². Note the pacing: the ball spends 2 s climbing the first 20 m and only 3 s covering the 45 m back to the ground, because it starts the descent from rest.
The symmetry of a vertical throw
When a body returns to the same level it was launched from, the flight is perfectly symmetric:
time up = time down, and it arrives back with exactly the launch speed reversed. This is why
T = 2u/g and why the fastest route to the height is often
H = gT²/8.
The symmetry breaks the moment the levels differ
A ball thrown up from a rooftop does not take equal times up and down — it has further to
fall than it climbed. In Example 2.3 the split is 2 s up, 3 s down. Only same-level throws are symmetric.
animated The three graphs for free fall from rest (NCERT Fig. 2.7). Acceleration is a flat line at −9.8 m s⁻²; velocity is a straight line through the origin; distance is a parabola. The sweep bar moves through all three together — each graph is the accumulated area of the one before it.
Two applications NCERT builds whole examples around
Stopping distance is in File 2. Here the two are Galileo’s law of odd numbers
(distances in successive equal intervals go as 1 : 3 : 5 : 7) and reaction time, which is measured
by dropping a ruler through a friend’s fingers. A 21.0 cm drop corresponds to about 0.2 s of
reaction time — a number worth carrying.
animated The ruler-drop timer. Because d = ½gt₀², a slow reaction lets far more ruler slip past: doubling the reaction time quadruples the distance caught.
Why this topic is Priority 1
Free fall appears in NEET both as a standalone numerical and as the setup for questions in Laws of
Motion and Work–Energy. It also carries an unusually high error rate — not because the physics
is hard, but because a single flipped sign silently produces one of the wrong options. Fixing your
convention habit here is worth more marks than learning any new formula.
2 · Formula sheet
Free fall from rest (down positive)
v = gt ; h = ½gt² ; v² = 2gh
Free fall (up positive, NCERT convention)
v = −9.8t ; y = −4.9t² ; v² = −19.6y
NCERT Example 2.4. Negative values are correct, not errors.
Thrown up, max height
H = u²/2g
∝ u².
Thrown up, time to top
t₁ = u/g
∝ u.
Thrown up, total flight time (same level)
T = 2u/g
Time up = time down.
Height from total flight time
H = gT²/8
Handy when only T is given.
Return speed at launch level
= u
Same magnitude, opposite direction.
Impact speed from height h
v = √(2gh)
Independent of mass.
Thrown from a height y₀
0 = y₀ + ut − ½gt²
Solve the quadratic; take the positive root.
Distance in the nth second of a fall
Sn = (g/2)(2n − 1)
Galileo’s odd numbers 1:3:5:7.
Distance in the last second
g/2 × (2T − 1)
T = total fall time.
Two bodies released τ apart
separation = gτ(t − τ/2)
Grows linearly — relative acceleration is zero.
Given height, given time
h ∝ g (same t) ; t ∝ 1/√g (same h)
Decide which one the question holds fixed.
3 · Exceptions and traps
Trap 1 — ‘dropped’ does not always mean from rest
A stone dropped from a rising balloon starts with the balloon’s upward velocity. A package
released from a moving aircraft keeps the aircraft’s velocity. ‘Dropped’ means zero
velocity relative to the carrier.
Trap 2 — discarding the second root
A given height below the peak is reached twice. Both positive roots of the quadratic are physical:
one going up, one coming down. Only negative roots get rejected.
Trap 3 — asymmetric launch and landing levels
Time up equals time down only for a same-level round trip. From a rooftop, the descent is longer.
Trap 4 — forgetting that g is constant only for small heights
NCERT is explicit: g can be taken as constant only if the height of fall is small compared with the
Earth’s radius. For satellite-scale problems this whole toolkit fails.
Trap 5 — assuming zero velocity means zero acceleration
At the top of a throw, v = 0 and a = −g simultaneously. Points to Ponder 4 says so directly, and
this appears as an assertion-reason item.
Exception — with air resistance, the symmetry dies
Real falling bodies experience drag. The ascent is then shorter than the descent, and the body
returns with less than its launch speed. Every result on this page assumes air resistance is neglected,
which is why NEET questions always say so explicitly. If a question mentions air resistance, do not use
these formulas.
4 · Numbers to remember
Fact
Value
g, standard value (NCERT)
9.8 m s⁻²
g, exam-friendly value when stated
10 m s⁻²
Fall distances at 1, 2, 3 s (g = 9.8)
4.9 m, 19.6 m, 44.1 m
Fall distances at 1, 2, 3 s (g = 10)
5 m, 20 m, 45 m
Drop-height → impact speed pairs (g = 10)
5 m→10, 20 m→20, 45 m→30, 80 m→40 m s⁻¹
NCERT Example 2.3 (20 m s⁻¹ from a 25 m roof)
rises 20 m; total time 5 s (2 s up, 3 s down)
NCERT Exercise 2.6 (29.4 m s⁻¹ throw)
44.1 m; returns in 6 s
NCERT Example 2.7 (ruler drop of 21.0 cm)
reaction time ≈ 0.2 s
NCERT Exercise 2.8 (drop from 90 m, g = 9.8)
first impact 42 m s⁻¹ at t = 4.3 s
Galileo’s odd numbers
1 : 3 : 5 : 7 : 9 : 11 …
5 · Scientists NEET names
Galileo Galilei (1564–1642)
The one name to know cold for this chapter. Two separate contributions are credited to him in NCERT:
He settled that acceleration should be defined as the rate of change of velocity with time
rather than with distance, after studying freely falling bodies and bodies on inclined planes. He found
that dv/dt is constant for all bodies in free fall, whereas the change of velocity
with distance decreases as the fall lengthens.
He established the law of odd numbers (Example 2.5, Table 2.2), and NCERT names him as the
first to make quantitative studies of free fall.
Remember the dates 1564–1642 — they occasionally appear in the option text itself.
6 · Twenty worked questions
Red left-border = previous-year question. Before starting each
one, write down your sign convention on the rough sheet. Every time.
Q1NEET-style
A ball is thrown vertically upwards with a velocity of 20 m s⁻¹ from the top of a multistorey building 25.0 m high. How high does the ball rise above the point of throw, and how long before it hits the ground? (g = 10 m s⁻²)
(a) 20 m, 4 s
(b) 25 m, 5 s
(c) 45 m, 3 s
(d) 20 m, 5 s
Given
v₀ = +20 m s⁻¹, launch height 25.0 m, a = −10 m s⁻²
Asked
rise above launch point, and total time of flight to the ground
Concept to use
Choose upward as positive with the origin at the ground, then treat the whole flight as one motion under constant acceleration. You do not need to split it at the top.
Formula or rule
v² = v₀² + 2a(y − y₀) ; y = y₀ + v₀t + ½at²
Baby steps
Rise: at the top v = 0, so 0 = 400 + 2(−10)(y − y₀), giving y − y₀ = 20 m.
Whole flight: 0 = 25 + 20t + ½(−10)t².
5t² − 20t − 25 = 0 ⇒ t² − 4t − 5 = 0.
(t − 5)(t + 1) = 0 ⇒ t = 5 s (reject t = −1 s).
Answer
It rises 20 m, and the total time to reach the ground is 5 s
Shortcut
The single-equation route is safer than splitting into up-leg and down-leg. NCERT itself says the second method is better because you need not worry about the path.
Trap
Splitting the flight and then forgetting that the ball falls 45 m on the way down, not 25 m. The 20 m it climbed has to be paid back too. This is NCERT Example 2.3.
Q2NEET-style
A player throws a ball upwards with an initial speed of 29.4 m s⁻¹. To what height does the ball rise, and after how long does it return to his hands? (g = 9.8 m s⁻², no air resistance)
(a) 44.1 m, 6 s
(b) 29.4 m, 3 s
(c) 44.1 m, 3 s
(d) 88.2 m, 6 s
Given
v₀ = 29.4 m s⁻¹, g = 9.8 m s⁻², returns to the launch level
Asked
maximum height and total time of flight
Concept to use
Symmetric flight: the ball returns to the same level it left, so time up = time down and the return speed equals the launch speed.
Formula or rule
H = v₀²/2g ; T = 2v₀/g
Baby steps
H = 29.4²/(2 × 9.8) = 864.36/19.6 = 44.1 m.
Time up: t = 29.4/9.8 = 3 s.
By symmetry the fall takes the same 3 s.
T = 6 s.
Answer
44.1 m, returning after 6 s
Shortcut
Notice 29.4 = 3 × 9.8, so the time up is exactly 3 s by inspection. Examiners choose multiples of 9.8 for precisely this reason — spot them.
Trap
Halving 6 s or doubling 44.1 m by mixing up which quantity the symmetry applies to. Time doubles for the round trip; height does not. This is NCERT Exercise 2.6(d).
Q3NEET-style
A ruler is dropped vertically through the gap between a person’s thumb and forefinger and is caught after falling 21.0 cm. The person’s reaction time is about (g = 9.8 m s⁻²)
(a) 0.2 s
(b) 0.1 s
(c) 0.4 s
(d) 0.6 s
Given
d = 21.0 cm = 0.21 m, released from rest, g = 9.8 m s⁻²
Asked
reaction time tr
Concept to use
The ruler is in free fall from rest, so the distance it falls before you catch it measures the time you took to react.
Formula or rule
d = ½gtr² ⇒ tr = √(2d/g)
Baby steps
tr = √(2 × 0.21 / 9.8).
= √(0.42/9.8) = √0.04286.
≈ 0.207 s ≈ 0.2 s.
Answer
about 0.2 s
Shortcut
Remember the landmark: 0.2 s of reaction → about 20 cm of ruler. It scales as the square root, so 0.4 s would need roughly 80 cm.
Trap
Working in centimetres and forgetting to convert. Using d = 21 gives 2.07 s, which would be a comically slow reaction. This is NCERT Example 2.7.
Q4PYQ · AIPMT 2013
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 s, the next 5 s and the next 5 s respectively. The relation between them is
(a) h₁ = h₂ = h₃
(b) h₁ = 2h₂ = 3h₃
(c) h₁ = h₂/3 = h₃/5
(d) h₂ = 3h₁ and h₃ = 3h₂
Given
free fall from rest, three consecutive 5 s intervals
Asked
relation between h₁, h₂, h₃
Concept to use
Galileo’s law of odd numbers applies to any equal intervals, not just one-second ones. Distances in successive equal intervals go as 1 : 3 : 5.
Formula or rule
h ∝ t², so successive equal intervals give 1 : 3 : 5
Write 1 : 3 : 5 first, then translate into whichever algebraic form the options use. Option (d) says h₃/h₂ = 3, but 5/3 ≠ 3, so it fails immediately.
Trap
Applying 1 : 3 : 5 only to one-second intervals. The interval length is irrelevant as long as all three are equal and the body starts from rest.
Q5PYQ · AIPMT 2010
A ball is dropped from a high platform at t = 0, starting from rest. After 6 s another ball is thrown downwards from the same platform with speed v. The two balls meet at t = 18 s. The value of v is (g = 10 m s⁻²)
(a) 55 m s⁻¹
(b) 75 m s⁻¹
(c) 65 m s⁻¹
(d) 85 m s⁻¹
Given
ball 1: dropped at t = 0; ball 2: thrown down at t = 6 s with speed v; they meet at t = 18 s
Asked
the speed v
Concept to use
‘They meet’ means both have fallen the same distance from the platform. Careful with the clock: ball 2 has been falling for only 12 s when they meet.
Take downward as positive here so no minus signs appear at all. When every body in a problem moves downward, that is always the smarter convention.
Trap
Using t = 18 s for the second ball as well. Its own clock starts at t = 6 s, so it has 12 s of flight, not 18 s.
Q6NEET-style
A body falling freely from rest covers 45 m in the last second of its fall. The total height from which it fell is (g = 10 m s⁻²)
(a) 80 m
(b) 125 m
(c) 100 m
(d) 180 m
Given
distance in the last second = 45 m, free fall from rest, g = 10 m s⁻²
Asked
total height
Concept to use
Let the total fall time be T. The last second is the interval from T − 1 to T. Use the nth-second formula with n = T.
Formula or rule
Slast = (g/2)(2T − 1) ; h = ½gT²
Baby steps
45 = (10/2)(2T − 1) = 5(2T − 1).
2T − 1 = 9 ⇒ T = 5 s.
h = ½(10)(25) = 125 m.
Answer
125 m
Shortcut
The distance in the last second of a free fall from rest is 5(2T − 1) with g = 10. Solve for T in one step, then square it.
Trap
Interpreting ‘last second’ as ‘first second’ and writing 45 = 5 m. Also: assuming the last second is half the journey — it is not; here it is 45 m out of 125 m.
Q7NEET-style
A stone is dropped from a balloon that is rising vertically with a speed of 10 m s⁻¹ when it is at a height of 75 m above the ground. The time taken by the stone to reach the ground is (g = 10 m s⁻²)
(a) 3 s
(b) 4 s
(c) 6 s
(d) 5 s
Given
balloon rising at 10 m s⁻¹ at height 75 m; stone released
Asked
time to reach the ground
Concept to use
A released object keeps the velocity of its carrier at the instant of release. The stone therefore starts moving upward at 10 m s⁻¹, not from rest.
Formula or rule
y = y₀ + v₀t + ½at² with upward positive
Baby steps
Take up as positive: v₀ = +10 m s⁻¹, a = −10 m s⁻², displacement = −75 m.
−75 = 10t − 5t².
5t² − 10t − 75 = 0 ⇒ t² − 2t − 15 = 0.
(t − 5)(t + 3) = 0 ⇒ t = 5 s.
Answer
5 s
Shortcut
Cross-check with the physics: if it had simply been dropped from rest, t = √(150/10) = 3.87 s. The upward toss must make it longer, and 5 s > 3.87 s. ✓
Trap
Treating the stone as starting from rest, which gives 3.87 s. ‘Dropped’ means zero velocity relative to the balloon, not relative to the ground.
Q8NEET-style
A ball is thrown vertically upward with a speed of 40 m s⁻¹. At what times is it at a height of 60 m above the point of projection? (g = 10 m s⁻²)
(a) only at t = 2 s
(b) only at t = 6 s
(c) at t = 2 s and t = 6 s
(d) at t = 3 s and t = 5 s
Given
v₀ = 40 m s⁻¹, y = 60 m, g = 10 m s⁻²
Asked
the time(s) at which y = 60 m
Concept to use
A given height below the maximum is reached twice: once going up and once coming down. The quadratic gives both roots and both are physical.
Formula or rule
y = v₀t − ½gt²
Baby steps
60 = 40t − 5t².
5t² − 40t + 60 = 0 ⇒ t² − 8t + 12 = 0.
(t − 2)(t − 6) = 0.
t = 2 s (rising) and t = 6 s (falling).
Answer
at t = 2 s and again at t = 6 s
Shortcut
The two roots straddle the time of maximum height, t = v₀/g = 4 s, symmetrically: 4 − 2 and 4 + 2. Find the peak time first and the roots come free.
Trap
Discarding the second root as ‘unphysical’. A negative root would be unphysical; a second positive root is the ball on its way back down. Max height here is 80 m, so 60 m is genuinely reached twice.
Q9NEET-style
Two identical balls are thrown vertically upward with initial speeds u and 2u. The ratio of the maximum heights they reach and of their times of flight are respectively
(a) 1:2 and 1:2
(b) 1:2 and 1:4
(c) 1:4 and 1:2
(d) 1:4 and 1:4
Given
initial speeds u and 2u, same g
Asked
ratio of maximum heights and ratio of times of flight
Concept to use
Height depends on the square of the launch speed; time depends on the first power. This mismatch is the whole question.
Formula or rule
H = u²/2g ; T = 2u/g
Baby steps
H ∝ u², so H₁:H₂ = u²:4u² = 1:4.
T ∝ u, so T₁:T₂ = u:2u = 1:2.
Answer: 1:4 and 1:2.
Answer
1:4 for heights and 1:2 for times
Shortcut
Squared quantity → height and stopping distance. Linear quantity → time. The same pair of rules governs the braking-distance questions in File 2.
Trap
Applying the same ratio to both. Papers rely on candidates carrying 1:4 across to the time as well.
Q10NEET-style
A ball is thrown vertically upward and returns to the thrower’s hand after 6 s. The maximum height reached is (g = 10 m s⁻²)
(a) 45 m
(b) 30 m
(c) 60 m
(d) 90 m
Given
total time of flight T = 6 s, returns to the same level
Asked
maximum height
Concept to use
For a symmetric flight, half the total time is the time to reach the top. Use that half-time in the free-fall distance formula.
Formula or rule
H = ½g(T/2)²
Baby steps
Time up = 6/2 = 3 s.
The descent from the top is a free fall from rest for 3 s.
H = ½(10)(9) = 45 m.
Check: u = gT/2 = 30 m s⁻¹, H = 30²/20 = 45 m. ✓
Answer
45 m
Shortcut
H = gT²/8 for a symmetric round trip of total time T. Here 10(36)/8 = 45 m in one line.
Trap
Using the full 6 s and getting 180 m. The 6 s covers the round trip; only 3 s of it is the climb.
Q11NEET-style
A body is dropped from rest from a height h. The ratio of the time taken to fall the first h/2 to the time taken to fall the second h/2 is
(a) 1 : 1
(b) 1 : (√2 − 1)
(c) (√2 − 1) : 1
(d) 1 : √2
Given
free fall from rest through height h, split into two equal halves
Asked
tfirst half : tsecond half
Concept to use
From rest, distance grows as t², so time grows as √(distance). The first half therefore eats most of the clock.
Formula or rule
t ∝ √h
Baby steps
Total time T = √(2h/g).
Time for the first h/2: t₁ = √(h/g) = T/√2.
t₂ = T − T/√2 = T(1 − 1/√2).
t₁:t₂ = 1 : (√2 − 1) ≈ 1 : 0.41.
Answer
1 : (√2 − 1)
Shortcut
About 71% of the total time is spent on the top half of the fall. If an option suggests the lower half takes longer, it is wrong on sight.
Trap
Answering 1 : 1. Equal distances take equal times only at constant speed, and a falling body is anything but.
Q12NEET-style
A ball is dropped from a height of 20 m. The speed with which it strikes the ground is (g = 10 m s⁻²)
(a) 20 m s⁻¹
(b) 10 m s⁻¹
(c) 40 m s⁻¹
(d) 200 m s⁻¹
Given
h = 20 m, dropped from rest, g = 10 m s⁻²
Asked
impact speed
Concept to use
Use the equation without time in it. Impact speed depends only on the drop height, not on the mass of the ball.
Formula or rule
v = √(2gh)
Baby steps
v² = 2(10)(20) = 400.
v = 20 m s⁻¹.
Time of fall, if needed: t = v/g = 2 s.
Answer
20 m s⁻¹
Shortcut
Memorise the pairs for g = 10: 5 m → 10 m s⁻¹, 20 m → 20 m s⁻¹, 45 m → 30 m s⁻¹, 80 m → 40 m s⁻¹. They recur constantly.
Trap
Reporting 200 m s⁻¹ by forgetting the square root. Option (d) is exactly v², planted for that slip.
Q13NEET-style
A stone is thrown vertically downward with an initial speed of 5 m s⁻¹ from a height of 30 m. The time it takes to reach the ground is (g = 10 m s⁻²)
(a) 2 s
(b) 1.5 s
(c) 2.45 s
(d) 3 s
Given
u = 5 m s⁻¹ downward, h = 30 m, g = 10 m s⁻²
Asked
time of fall
Concept to use
Take downward as positive so every quantity is positive and no sign errors are possible.
Formula or rule
h = ut + ½gt²
Baby steps
30 = 5t + 5t².
t² + t − 6 = 0.
(t + 3)(t − 2) = 0.
t = 2 s (reject t = −3 s).
Answer
2 s
Shortcut
Compare with a plain drop: √(60/10) = 2.45 s. The downward throw must beat that, and 2 s does. Use this to eliminate options (c) and (d).
Trap
Writing 30 = 5t − 5t² by mixing conventions — taking u as positive downward but g as negative. Pick one direction and stay in it.
Q14NEET-style
For a body falling freely from rest, the ratio of its velocities at the ends of the 1st, 2nd and 3rd seconds is
(a) 1 : 3 : 5
(b) 1 : 2 : 3
(c) 1 : 4 : 9
(d) 1 : √2 : √3
Given
free fall from rest
Asked
ratio of velocities at t = 1, 2, 3 s
Concept to use
Velocity grows linearly with time under constant acceleration; distance grows as the square. The question asks about velocity.
Formula or rule
v = gt
Baby steps
v₁ = g, v₂ = 2g, v₃ = 3g.
Ratio = 1 : 2 : 3.
(For contrast: total distances go 1 : 4 : 9, and distances in each second go 1 : 3 : 5.)
Answer
1 : 2 : 3
Shortcut
Three different ratios live in this one situation. Tag them: velocity 1:2:3, total distance 1:4:9, distance per second 1:3:5.
Trap
Reaching for 1:3:5 out of habit because Galileo’s law is fresh in mind. That ratio belongs to distances in successive intervals, not to velocities.
Q15NEET-style
A ball is dropped from a height of 90 m onto a floor. At each collision it loses one tenth of its speed. The speed with which it first strikes the floor, and the time taken to do so, are (g = 9.8 m s⁻²)
(a) 42 m s⁻¹, 8.6 s
(b) 37.8 m s⁻¹, 4.3 s
(c) 44.1 m s⁻¹, 4.5 s
(d) 42 m s⁻¹, 4.3 s
Given
h = 90 m, dropped from rest, g = 9.8 m s⁻²
Asked
first impact speed and the time to first impact
Concept to use
The one-tenth loss happens after the first impact, so it does not affect the first descent at all. Read the sequence of events carefully.
Formula or rule
v = √(2gh) ; t = v/g
Baby steps
v² = 2(9.8)(90) = 1764.
v = 42 m s⁻¹ exactly.
t = 42/9.8 = 4.29 s ≈ 4.3 s.
(It rebounds at 0.9 × 42 = 37.8 m s⁻¹ — that is the next stage, and it is what option (b) is testing.)
Answer
42 m s⁻¹ after 4.3 s
Shortcut
1764 is a perfect square (42²). Examiners choose 90 m with g = 9.8 precisely so the arithmetic lands clean — if your answer is untidy, re-check the multiplication.
Trap
Applying the 10% loss before the first impact and answering 37.8 m s⁻¹. This is the setup of NCERT Exercise 2.8, where the speed–time graph is a sawtooth with each peak 90% of the previous one.
Q16NEET-style
A particle is thrown vertically upward. At the highest point of its motion,
(a) both its velocity and acceleration are zero
(b) its velocity is zero and its acceleration is zero only momentarily
(c) its acceleration is zero and its velocity is g
(d) its velocity is zero and its acceleration is g downward
Given
particle at the top of a vertical throw
Asked
the correct statement about v and a there
Concept to use
Zero velocity at an instant does not imply zero acceleration. The velocity is passing through zero; the acceleration is what makes it pass through.
Formula or rule
a = −g throughout the flight, whatever v is doing
Baby steps
At the top the particle is momentarily at rest, so v = 0.
Gravity does not switch off — it is still pulling downward with a = g ≈ 9.8 m s⁻².
If a really were zero there, the particle would hang in the air forever.
Hence (d).
Answer
Its velocity is zero and its acceleration is g directed downward
Shortcut
NCERT Points to Ponder 4 states this outright: a particle may be momentarily at rest and still have non-zero acceleration. Quote it if the question is assertion-reason.
Trap
Reasoning that ‘nothing is moving, so nothing is accelerating’. Acceleration describes the change in velocity, and the velocity is changing fastest in sign exactly at that instant.
Q17NEET-style
An object is dropped from rest on the Moon, where the acceleration due to gravity is one sixth of that on Earth. If it falls a distance h on Earth in time t, then in the same time t on the Moon it falls
(a) 6h
(b) h
(c) h/6
(d) h/36
Given
gmoon = gearth/6, same time t, both from rest
Asked
distance fallen on the Moon
Concept to use
With the time held fixed, distance is directly proportional to g. (If the distance had been held fixed instead, the time would scale as √6.)
Formula or rule
h = ½gt²
Baby steps
hearth = ½g t².
hmoon = ½(g/6)t².
Ratio = 1/6, so the Moon fall is h/6.
Answer
h/6
Shortcut
Ask yourself which quantity the question is holding constant — time or distance. Same time → h ∝ g. Same height → t ∝ 1/√g.
Trap
Switching to the same-height comparison and answering with √6. Read the question’s fixed quantity before choosing the proportionality.
Q18NEET-style
Galileo’s law of odd numbers states that the distances traversed during equal intervals of time by a body falling from rest are in the ratio
(a) 1 : 2 : 3 : 4
(b) 1 : 3 : 5 : 7
(c) 1 : 4 : 9 : 16
(d) 2 : 4 : 6 : 8
Given
body falling from rest, equal successive time intervals τ
Asked
the ratio of distances traversed in successive intervals
Concept to use
The totals go as the squares; the differences between consecutive squares are the odd numbers.
Formula or rule
y = ½gτ² × n², differences give (2n − 1)
Baby steps
Positions after 0, τ, 2τ, 3τ, 4τ: proportional to 0, 1, 4, 9, 16.
Successive differences: 1, 3, 5, 7.
Hence the ratio 1 : 3 : 5 : 7 …
Answer
1 : 3 : 5 : 7 …
Shortcut
Option (c) is the ratio of total distances and option (a) is the ratio of velocities. All three ratios come from the same table — know which column each one is.
Trap
Attributing the law to Newton. NCERT names Galileo Galilei (1564–1642) as the first to make quantitative studies of free fall, and Table 2.2 is the proof.
Q19NEET-style
Two balls are dropped from the top of a tower, the second one exactly 1 s after the first. The separation between them 2 s after the first ball is released is (g = 10 m s⁻²)
(a) 5 m
(b) 10 m
(c) 20 m
(d) 15 m
Given
ball 1 released at t = 0, ball 2 at t = 1 s, both from rest; separation asked at t = 2 s
Asked
distance between the two balls
Concept to use
Compute each ball’s own fall using its own elapsed time, then subtract. The gap keeps widening because the leading ball is always faster.
Formula or rule
s = ½gt²
Baby steps
Ball 1 has fallen for 2 s: s₁ = ½(10)(4) = 20 m.
Ball 2 has fallen for 1 s: s₂ = ½(10)(1) = 5 m.
Separation = 20 − 5 = 15 m.
Answer
15 m
Shortcut
The separation after the second release grows as gτ(t − τ/2), i.e. it increases linearly with time — their relative acceleration is zero because both feel the same g.
Trap
Assuming the gap stays fixed at the 5 m the second ball ‘lost’. Relative velocity between them grows at gτ = 10 m s⁻¹ per second even though relative acceleration is zero.
Q20NEET-style
A body in free fall near the Earth’s surface is said to be in free fall only if
(a) it starts from rest
(b) it falls from a height comparable to the Earth’s radius
(c) air resistance is neglected and only gravity acts
(d) its mass is small
Given
NCERT definition of free fall (Example 2.4)
Asked
the defining condition
Concept to use
Free fall is defined by the forces acting, not by the initial velocity. A ball thrown upward is in free fall the moment it leaves the hand.
Formula or rule
Definition: motion under gravity alone, with air resistance neglected
Baby steps
NCERT: an object released near the Earth’s surface is accelerated downward under gravity; if air resistance is neglected it is said to be in free fall.
The starting velocity is irrelevant — upward, downward or zero, it is still free fall.
g may be treated as constant only if the fall height is small compared with the Earth’s radius, which rules out (b).
Mass does not appear anywhere in the kinematics, ruling out (d).
Answer
Air resistance is neglected and only gravity acts
Shortcut
Free fall is a statement about the force, not the motion. That single reframing kills most conceptual questions on this definition.
Trap
Thinking a body thrown upward is not in free fall. It is — its acceleration is −g throughout, including at the top.
Answer key
dQ1
aQ2
aQ3
cQ4
bQ5
bQ6
dQ7
cQ8
cQ9
aQ10
bQ11
aQ12
aQ13
bQ14
dQ15
dQ16
cQ17
bQ18
dQ19
cQ20
NEET 2027 preparation — built for Aamirah Fathima. Source: NCERT Physics Class XI,
Chapter 2 (Reprint 2026–27). Single self-contained file — print directly from any browser.