For a projectile launched with speed v₀ at angle θ₀ above the horizontal, with origin at launch point and y-axis upward: x(t) = (v₀ cos θ₀) t, y(t) = (v₀ sin θ₀) t − ½ g t². Trajectory: y = (tan θ₀) x − [g / (2 v₀² cos² θ₀)] x² (a parabola).
-- NCERT Class 11 Physics, Ch. 3, p. 39Projectile Motion
8 MCQs6 revision cards9-step worked example
Source: NCERT KinematicsPYQ coverage: NEET 2021, 2022Official key: NTA-verifiedLast updated: 21 Sep 2026
Projectile Motion, explained for NEET
Projectile motion is the motion of an object launched into the air with some initial velocity and then subject only to gravitational acceleration — no air resistance, no engine thrust. NCERT Class 11 Physics Chapter 3 (page 39) establishes the key principle: treat horizontal and vertical motions independently. Horizontally, acceleration is zero and the component v₀ cos θ stays constant throughout the flight. Vertically, the object undergoes free fall with acceleration g downward.
The trap that costs marks: confusing the maximum-height formula with the range formula. Both contain v₀ and θ, but the structures differ. Maximum height uses sin²θ and divides by 2g. Range uses sin(2θ) and divides by g alone. Plugging into the wrong formula is a common distractor source in NEET — the numbers look plausible but are off by a factor involving sin vs sin².
A second high-frequency trap involves the angle reference. When a problem states "launched at angle θ with the vertical," the horizontal component is v₀ sin θ (not v₀ cos θ). The two conventions are complementary — θ from horizontal and θ from vertical sum to 90° — but under exam pressure, students default to cos without checking the reference axis.
At the highest point, the vertical component of velocity is zero. The speed at the apex equals the horizontal component alone: v₀ cos θ (when θ is measured from the horizontal). A distractor claiming speed = 0 at the top is true only for purely vertical launch (θ = 90°).
The standard formulas — H = v₀² sin²θ/(2g) and R = v₀² sin(2θ)/g — assume launch and landing at the same height, negligible air resistance, and constant g. For problems involving cliffs or non-level ground, use the full component-wise kinematic equations instead.
A cross-topic trap: a particle exits uniform circular motion (radius R, period T, speed v = 2πR/T) and is launched as a projectile. The UCM speed feeds directly into projectile formulas. Do not confuse the radius R with the maximum height H.
Can you answer these Projectile Motion MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In projectile motion (neglecting air resistance), which quantity remains constant throughout the flight?
Show answer and why every option is right or wrong
Answer: C. In projectile motion with no air resistance, the horizontal acceleration is zero (NCERT Class 11 Physics Chapter 3, page 39). Therefore the horizontal component of velocity v₀ cos θ is conserved throughout the trajectory.
Why A is wrong: A is wrong because the vertical component changes continuously due to gravitational acceleration g acting downward.
Why B is wrong: B is wrong because speed is the magnitude of the total velocity vector. Since the vertical component changes, the speed changes — it is minimum at the highest point (equal to v₀ cos θ) and maximum at launch/landing.
Why D is wrong: D is wrong because horizontal acceleration is zero in ideal projectile motion, not g. The acceleration g acts only in the vertical direction.
MCQ 2Easy RecallPractice
The range formula R = v₀² sin(2θ)/g gives maximum range when the launch angle θ (measured from the horizontal) is:
Show answer and why every option is right or wrong
Answer: D. R is proportional to sin(2θ). sin(2θ) is maximised when 2θ = 90°, i.e. θ = 45° (NCERT Class 11 Physics Chapter 3, page 40).
Why A is wrong: A is wrong because sin(2 × 30°) = sin 60° = √3/2 ≈ 0.866, which is less than 1.
Why B is wrong: B is wrong because sin(2 × 60°) = sin 120° = √3/2 ≈ 0.866, the same as for 30° (complementary angles give equal range but not maximum range).
Why C is wrong: C is wrong because sin(2 × 90°) = sin 180° = 0, giving zero range — this is purely vertical motion.
MCQ 3Easy RecallPractice
The standard projectile range formula R = v₀² sin(2θ)/g is valid only when:
Show answer and why every option is right or wrong
Answer: A. The derivation of R = v₀² sin(2θ)/g assumes same-level launch and landing, no air drag, and uniform g over the trajectory (NCERT Class 11 Physics Chapter 3, page 40). Violating any assumption invalidates the formula.
Why B is wrong: B is wrong because asymmetric launch/landing heights require full kinematic decomposition — the simple range formula does not apply (trap: mistake: projectile air support).
Why C is wrong: C is wrong because the formula explicitly neglects air resistance. Including drag changes the trajectory from a parabola and invalidates this closed-form expression (trap: mistake: projectile air support).
Why D is wrong: D is wrong because at θ = 0° the projectile is launched horizontally with zero vertical component, giving zero range by this formula (sin(0) = 0). The formula works for any angle between 0° and 90°.
MCQ 4Direct ApplicationPractice
A projectile is launched at 20 m/s at an angle of 30° above the horizontal. Taking g = 10 m/s², the maximum height attained is:
Show answer and why every option is right or wrong
Answer: B. H = v₀² sin²θ / (2g) = (20)² × sin²(30°) / (2 × 10) = 400 × 0.25 / 20 = 5 m. The key is using sin²θ (not sin θ or sin 2θ) and the factor 2g in the denominator.
Why A is wrong: A is wrong — this result (20 m) comes from using v₀²/(2g) = 400/20 = 20, which omits the sin²θ factor entirely. That formula gives max height only for vertical launch (θ = 90°) (trap: projectile height vs range formula).
Why C is wrong: C is wrong — this result (10 m) comes from using v₀² sin θ / (2g) = 400 × 0.5 / 20 = 10, using sin θ instead of sin²θ (trap: projectile height vs range formula).
Why D is wrong: D is wrong — this could arise from an arithmetic error such as using sin²(30°) = 1/8 or misplacing a factor of 2.
MCQ 5Direct ApplicationPractice
A projectile is launched with speed v₀ at 60° to the vertical. Its speed at the highest point of the trajectory is:
Show answer and why every option is right or wrong
Answer: C. C is correct. The angle is measured from the vertical, so the angle above the horizontal is 90° − 60° = 30°. Only the horizontal component survives at the apex, and it is constant throughout the flight: v₀ cos 30° = v₀√3/2. Equivalently, taking components directly from the vertical, the horizontal part is v₀ sin 60°, which is the same number. (trap: projectile horizontal component angle reference)
Why A is wrong: A is wrong because the speed is zero at the apex only for a launch straight up. Here a horizontal component exists at launch, nothing acts horizontally to change it, and so it is still there at the top.
Why B is wrong: B is wrong because v₀/2 is v₀ cos 60°, which would be the horizontal component if the 60° were measured from the horizontal. It is measured from the vertical, so 60° is the angle whose SINE gives the horizontal part. (trap: projectile horizontal component angle reference)
Why D is wrong: D is wrong because v₀ is the launch speed. Only the horizontal component is unchanged; the vertical component has been removed by gravity by the time the projectile reaches the top, so the speed there is smaller than v₀.
MCQ 6Direct ApplicationPractice
A ball is launched at 40 m/s at 45° above the horizontal. Taking g = 10 m/s², the horizontal range is:
Show answer and why every option is right or wrong
Answer: B. R = v₀² sin(2θ)/g = (40)² × sin(90°)/10 = 1600 × 1/10 = 160 m. At θ = 45°, sin(2θ) = sin 90° = 1, giving maximum range for this speed.
Why A is wrong: A is wrong — 80 m comes from using v₀² sin θ / g = 1600 × (√2/2)/10 ≈ 113 m (not 80), or from halving the correct answer. A common route to 80 is using v₀² sin²θ / g = 1600 × 0.5 / 10 = 80, which is the height formula's numerator divided by g instead of 2g (trap: projectile height vs range formula).
Why C is wrong: C is wrong — 320 m comes from doubling the range, possibly by using 2v₀² sin(2θ)/g, which has an extra factor of 2 with no physical basis.
Why D is wrong: D is wrong — 40 m equals the launch speed numerically but has no connection to the range formula.
MCQ 7CalculationPractice
A particle moves in a circle of radius 2.0 m with a period of 2.0 s. It is then launched vertically upward with the same speed it had during the circular motion. Taking g = 10 m/s² and π² ≈ 10, the maximum height reached is:
Show answer and why every option is right or wrong
Answer: C. C is correct. The speed in uniform circular motion is v = 2πR/T = 2π(2.0 m)/(2.0 s) = 2π m/s, so v² = 4π² ≈ 40 m²/s². Launched straight up with that speed, the particle rises until all of its kinetic energy has become potential energy: H = v²/(2g) = 40/(2 × 10) = 2.0 m.
Why A is wrong: A is wrong because 1.0 m comes from dividing by 4g instead of 2g — an extra factor of 2 in the denominator of H = v²/(2g).
Why B is wrong: B is wrong because 4.0 m comes from H = v²/g, dropping the factor of 2 altogether.
Why D is wrong: D is wrong because 0.5 m follows from v = πR/T, which forgets the 2 in the circumference 2πR. That halves the speed, and since H depends on v² it quarters the height.
MCQ 8CalculationPractice
Two projectiles are launched from the same point with the same speed v₀. Projectile A is launched at 30° and projectile B at 60° above the horizontal. Neglecting air resistance, the ratio of their maximum heights H_A : H_B is:
Show answer and why every option is right or wrong
Answer: A. H = v₀² sin²θ/(2g). H_A/H_B = sin²(30°)/sin²(60°) = (1/2)²/(√3/2)² = (1/4)/(3/4) = 1/3. So H_A : H_B = 1 : 3.
Why B is wrong: B is wrong — complementary angles (30° and 60°) give equal RANGE, not equal maximum height. The range formula has sin(2θ) which is the same for both (sin 60° = sin 120°), but the height formula has sin²θ, which differs (trap: projectile height vs range formula).
Why C is wrong: C is wrong (3:1) — this is the inverted ratio, arising from swapping A and B in the calculation or computing H_B/H_A instead of H_A/H_B.
Why D is wrong: D is wrong (1:√3) — this would come from using sinθ instead of sin²θ in the height formula: sin 30°/sin 60° = (1/2)/(√3/2) = 1/√3 (trap: projectile height vs range formula).
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Projectile Motion: quick recall before you leave
How do you solve a Projectile Motion question? A worked example
Pattern: NEET pattern: projectile max height from launch params (direct_application, anchored to PYQ 2023 F3 Q35)
- 1
Given
A projectile is launched with speed v₀ = 50 m/s at angle θ = 30° above the horizontal. Take g = 10 m/s² (exact, problem-defined).
- 2
Required
Find the maximum height H attained above the launch point.
- 3
Concept
In projectile motion with no air resistance, the vertical and horizontal motions are independent. The maximum height is reached when the vertical component of velocity becomes zero. The formula relating launch parameters to maximum height is derived from the kinematic equation v² = u² + 2as applied to the vertical direction (NCERT Class 11 Physics Chapter 3, page 40).
- 4
Formula
H = v₀² sin²θ / (2g)
- 5
Substitution
H = (50)² × sin²(30°) / (2 × 10)
H = 2500 × (0.5)² / 20
H = 2500 × 0.25 / 20 - 6
Calculation
H = 625 / 20 = 31.25 m
Note on exact constants: g = 10 m/s² is an exact problem-defined value. The integer 2 in the denominator and sin 30° = 0.5 (exact trigonometric value) are mathematical constants. These do not limit the precision of the answer. The answer's precision is governed by v₀ = 50 m/s (2 significant figures), giving H = 31 m to 2 significant figures. - 7
Final answer
H = 31.25 m (or 31 m to 2 significant figures, matching the precision of v₀).
- 8
Common trap
The most common error is using the range formula R = v₀² sin(2θ)/g instead of the height formula. That gives R = 2500 × sin(60°)/10 = 2500 × 0.866/10 ≈ 216.5 m — a plausible-looking number that answers the wrong question entirely. The telltale difference: height has sin² and 2g; range has sin(2θ) and g.
- 9
Similar NEET-style question
A ball is thrown at 40 m/s at 60° above the horizontal. Find the maximum height. (Answer: H = (40)² × sin²(60°)/(2 × 10) = 1600 × 0.75/20 = 60 m.)
---
What to remember before solving Projectile Motion questions
Range R = v₀² sin(2θ₀) / g — maximum at θ₀ = 45°. Maximum height H = v₀² sin² θ₀ / (2g). Time of flight T_f = 2 v₀ sin θ₀ / g. (All neglect air resistance and treat g as constant.)
-- NCERT Class 11 Physics, Ch. 3, p. 40Worked Example
Example 3.8 — Projectile range and height
A cricket ball thrown at 28 m/s at 30° above horizontal: maximum height H = 28² sin²(30°) / (2 · 9.8) ≈ 10.0 m; time to return to same level T_f = 2(28)sin(30°)/9.8 ≈ 2.86 s; horizontal range R = 28² sin(60°) / 9.8 ≈ 69.3 m.
-- NCERT Class 11 Physics, Ch. 3, p. 40Which Projectile Motion formulas do you need for NEET?
2 formulas — click to collapse
Projectile maximum height
Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H | Maximum height (above launch) | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle | rad/deg |
| g | Gravitational acceleration | m/s^2 |
Valid when
- Air resistance neglected
- Constant g over trajectory
Projectile horizontal range
For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | Horizontal range | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle above horizontal | rad (or deg with sin in deg) |
| g | Gravitational acceleration | m/s^2 |
Valid when
- Launch and landing are at the same vertical height
- Air resistance neglected
- g treated as constant over the trajectory
Do NOT use when
- Launch and landing heights differ (use full kinematics)
- Significant air drag (e.g. table-tennis ball, badminton shuttle)
- Variation of g (ballistic trajectories spanning large altitude changes)
Where do students lose marks on Projectile Motion?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
4 items — click to collapse
Category: Similar Terms
Student plugs into v₀² sin(2θ)/g (range) when asked for maximum height, or vice versa. The two share v₀ and θ but have different sin-vs-sin² and 2g-vs-g terms.
When it triggers
Question mentions launch speed and angle and asks for max height (H) or range (R). Distractors include the wrong formula's answer.
How to avoid
Memorise BOTH formulas explicitly: H = v₀² sin² θ / (2g) (note sin²); R = v₀² sin(2θ) / g (note sin of doubled angle). Check by setting θ = 45°: max range, half max height.
Category: Sign Convention
Student plugs angle θ into v cos θ when the question states 'angle with the vertical' (which makes the horizontal component v sin θ).
When it triggers
Question phrases like 'thrown at angle θ with the vertical direction' or 'with horizontal'.
How to avoid
Always identify reference axis explicitly. From horizontal: vx = v cos θ, vy = v sin θ. From vertical: vx = v sin θ, vy = v cos θ. The two are complementary (θ_h + θ_v = 90°).
Category: Overthinking
Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.
When it triggers
Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.
How to avoid
Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.
Root cause: concept gap
Correction
Standard range R = v0^2 sin(2*theta)/g and H = v0^2 sin^2(theta)/(2g) assume (i) launch and landing at the same height, (ii) negligible air drag, and (iii) constant g. For asymmetric trajectories, use the full kinematic decomposition along x and y.
Wrong option pattern
Distractor applies R = v0^2 sin(2*theta)/g to a projectile launched from a cliff.
More in Kinematics: 10 exam traps and mistakes · 4 formulas · 7 question patterns from its other lessons.
Projectile Motion questions from past NEET papers
2 questions from NEET 2021, 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
110 ms⁻¹
2Zero
35√3 ms⁻¹
45 ms⁻¹
NTA Answer: Option 3(final)
NEET 2021
1θ = sin⁻¹[(2gT²)/(π²R)]^(1/2)
2θ = cos⁻¹[(gT²)/(π²R)]^(1/2)
3θ = cos⁻¹[(π²R)/(gT²)]^(1/2)
4θ = sin⁻¹[(π²R)/(gT²)]^(1/2)
NTA Answer: Option 1(final)
How does NEET ask about Projectile Motion?
3 recurring patterns from past papers — click to collapse
A particle in uniform circular motion (period T, radius R) is subsequently launched vertically upward with the SAME speed (v = 2πR/T); find max height. H = v²/(2g) = (2πR/T)²/(2g).
Multi StepMedium
Common distractors
uses radius as height
Surface confusion of geometric R and projectile H
Given launch speed v₀ and angle θ above horizontal, find maximum height. H = v₀² sin² θ / (2g). Common values plug in cleanly when sin θ ∈ {0.5, 0.707, 0.866}. Distractors test (i) using sin instead of sin², (ii) forgetting the factor of 2 in denominator, (iii) using g = 10 vs g = 9.8 inconsistently.
Direct ApplicationEasy
Common distractors
uses sin instead of sin squared
Confusing the height formula H = v²sin²θ/(2g) with the range 2v sin θ / g
Projectile launched at angle θ; find SPEED at highest point. The vertical component vanishes at the apex; horizontal component v₀ cos θ is conserved. Common gotcha: 'angle with the vertical' vs 'angle with the horizontal'.
Direct ApplicationEasy
Common distractors
uses sin instead of cos
Sign-of-angle-axis confusion
speed zero at top
True only for vertical motion (no horizontal v₀)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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