Horizontal Projectile Motion

NEET > Physics > Kinematics > Motion In Two Dimension > Horizontal Projectile Motion

Unit Progress

NEET Physics — Motion In Two Dimension

Horizontal Projectile Motion – Complete Notes, Revision, Important Questions & Downloads

Horizontal projectile motion is the special case where a body is projected with a purely horizontal initial velocity u from a height H above the ground (θ = 0°). Three subtopics are covered: Trajectory of Horizontal Projectile (y = gx²/2u²), Time of Flight in Horizontal Projection (T = √(2H/g)), and Horizontal Range in Horizontal Projection (R = u√(2H/g)). NEET tests this by asking how far from the cliff base a horizontally launched body lands, the speed of impact, or the angle of velocity at impact. A classic multi-body scenario ('three particles from same height — same or different time?') also appears directly in NEET.

⬇ Download Notes PDF View Important Questions →

7 Subtopicsθ = 0° CaseT = √(2H/g)

Expected QuestionsQ

0–1

Horizontal projection appears in NEET as the cliff-drop scenario or as a component of multi-step energy/2D questions approximately once every 2 years.

Time Required⏱

1.5 hrs

20 min for trajectory derivation and parabola shape; 20 min for T and R formulas; 30 min on the three-particle equal-time result; 20 min on velocity and angle at impact. Keep focus on conceptual traps.

Difficulty⚡

Easy–Medium

The formulas are simpler than oblique projection (no angle component). Main difficulty: velocity at impact has two components and the angle requires trigonometry. The equal-time trap (all drop from same height in same time) is a high-yield conceptual trap.

NRI USA Curriculum GapUS

Low

AP Physics 1 fully covers horizontal projectile motion. The NEET-specific element is the three-particle simultaneous scenario and recognising that the trajectory equation y = gx²/(2u²) is a parabola through the origin.

7Subtopics

8+Practice Questions

4Free Downloads

1.5 hrsPrep Time

⬇ Get Free Downloads

NEET Weightage — Horizontal Projectile Motion

Motion In Two Dimension (Chapter 3)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
2022	1	1 Q	4
2021	0	0 Q	0
2020	1	1 Q	4
2019	0	0 Q	0
6-Year Total (2019–2024)	2–4		8–16

Equal time for all horizontal projections from the same height: body A projected horizontally at 10 m/s, body B at 20 m/s, body C dropped — all reach the ground simultaneously. The time T = √(2H/g) depends only on height.
Velocity at impact: v_x = u (horizontal, unchanged), v_y = gT = g√(2H/g) = √(2gH). Speed on impact = √(u² + 2gH). This formula connects to energy conservation: (1/2)mv² = (1/2)mu² + mgH.
Angle of velocity at impact: tan(α) = v_y/v_x = √(2gH)/u. NEET may give this angle and ask for u.

📊

0.5

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy–Medium

Difficulty

How to Prepare Horizontal Projection for NEET

Derive the trajectory equation once to understand the parabola Horizontal: x = ut → t = x/u. Vertical: y = (1/2)gt² = (1/2)g(x/u)² = gx²/(2u²). This is a parabola through the origin with the projectile moving along it from (0,0) downward. NEET may show a curve and ask which axis is x and which is y.

Memorise T = √(2H/g) as the defining formula Time of flight depends only on height H, not on horizontal speed u. This is the key concept: all bodies projected horizontally (at any speed) from the same height land at the same time — only horizontal range differs.

Calculate velocity at impact vectorially v_x = u (horizontal). v_y = gT = g√(2H/g) = √(2gH) (downward). Speed = √(u² + 2gH). Angle with horizontal: tan(α) = v_y/v_x. NEET gives two of these three and asks for the third.

Equal-time conceptual question Three particles A, B, C from same height H: A dropped, B horizontal at 10 m/s, C horizontal at 20 m/s. All reach ground at T = √(2H/g). NEET marks this as assertion: 'They take equal time.' Reason: 'Time depends only on vertical motion.' Both TRUE and reason correctly explains.

Study Materials — Horizontal Projectile Motion

PDF · Cheat Sheet · MCQ Set · PYQ

📘

Full Notes

All 3 subtopics: trajectory as parabola through origin, time of flight formula, horizontal range calculation, velocity at impact (magnitude and direction), equal-time result for three particles from same height.

7 subtopics6 pagesDiagram-heavy

Download Notes

📗

Formula Sheet

y = gx²/(2u²); T = √(2H/g); R = u√(2H/g); v_y = √(2gH); v = √(u² + 2gH); tan(α) = √(2gH)/u. Table: u, H → T, R, v_impact.

6 formulas1 pageTable format

Download Sheet

📙

MCQ Practice

15 questions: find T given H, find R given u and H, find impact velocity, identify trajectory shape, equal-time scenario, and compare ranges from same height at different horizontal speeds.

15 MCQsGraded difficultySolved

Download MCQs

📒

PYQ

Year-tagged NEET previous-year horizontal projection questions with full solutions — cliff problems, equal-time assertions, and impact velocity calculations.

8+ year-tagged Qs2015–2024Step-by-step solutions

Download PYQs

Subtopics in Horizontal Projectile Motion

2-Column Table

Column A	Column B
Trajectory of Horizontal Projectile↗	Time of Flight in Horizontal Projection↗
Horizontal Range in Horizontal Projection↗	Units : second↗
Uniform Circular Motion↗	Angular displacement↗
Angular velocity↗

Rapid Revision — Horizontal Projectile Motion

Concept → Trap → Example

1) Trajectory of Horizontal Projectile

Core

Initial velocity: u (horizontal only, θ = 0°). Horizontal: x = ut. Vertical: y = (1/2)gt². Eliminate t: y = gx²/(2u²). This is a parabola through the origin, opening downward.

The trajectory y = gx²/(2u²) shows: y ∝ x². On the x-y plane (x = horizontal distance, y = vertical drop), the path curves downward.
For larger u: same x corresponds to smaller y (the projectile is flatter). The parabola is wider for higher initial speed.
Important: the vertex of the parabola is at the launch point (0,0). The path never goes above the launch height.

Example (NEET-style)Ball launched horizontally at u = 5 m/s from height H = 20 m. Trajectory: y = 10x²/(2×25) = x²/5. At x = 5 m: y = 5 m (dropped 5 m). At x = 10 m: y = 20 m (at ground). This confirms R = 10 m for H = 20 m.

2) Time of Flight in Horizontal Projection

High Yield

Time of flight depends only on height H: T = √(2H/g). The horizontal velocity u does not affect T. All bodies projected horizontally from the same height land simultaneously — regardless of their horizontal speed.

Derivation: vertical motion from height H with u_y = 0 and a = g: H = (1/2)gT² → T = √(2H/g).
The three-particle result: A (dropped), B (horizontal 10 m/s), C (horizontal 20 m/s) from height H = 20 m all land at T = √(2×20/10) = 2 s simultaneously.
Trap: 'A ball projected horizontally at higher speed takes longer to fall' — FALSE. T = √(2H/g) is independent of u.

Example (NEET-style)From height H = 45 m: T = √(2×45/10) = √9 = 3 s. Any horizontal speed gives the same T = 3 s. Vertical velocity at ground: v_y = gT = 10×3 = 30 m/s downward.

3) Horizontal Range in Horizontal Projection

Core

Range R = u × T = u × √(2H/g) = u√(2H/g). Speed at impact: v = √(u² + v_y²) = √(u² + 2gH). Angle with horizontal: tan(α) = v_y/v_x = √(2gH)/u.

Range R ∝ u for fixed H: doubling horizontal speed doubles horizontal range.
Range R ∝ √H for fixed u: to double the range from same u, height must be quadrupled.
Speed at impact = √(u² + 2gH) — this equals the speed obtained by energy conservation, confirming that mechanical energy is conserved (gravity is conservative).

Example (NEET-style)u = 15 m/s, H = 80 m, g = 10 m/s². T = √(16) = 4 s. R = 15×4 = 60 m. v_y = 10×4 = 40 m/s. Speed at impact = √(225 + 1600) = √1825 ≈ 42.7 m/s. Angle: tan(α) = 40/15 = 2.67 → α ≈ 69.4°.

US Curriculum Gaps — Horizontal Projectile Motion

Topics in this section are tested in NEET but less emphasised in standard US physics courses.

Three-Particle Equal-Time Scenario (AP Physics 1 Gap)

AP Physics 1 teaches that time of flight for a horizontal projectile is independent of horizontal speed. However, the explicit three-particle scenario (A dropped, B and C thrown horizontally at different speeds — all from same height) is a direct NEET assertion-reason question format that AP Physics 1 does not drill as a distinct exam type. NEET students must immediately recognise 'same height → same time' as an assertion-reason pair.

AP Physics 1 covers the concept but does not test it as assertion-reason
NEET assertion: 'Three particles from same height take same time to land.' Reason: 'Time depends only on vertical free-fall.' Both statements true, reason is correct explanation.
This appears in NEET about once every 3 years — a reliable 4-mark question for students who drill it.

Trajectory Equation as a Parabola Through Origin (AP Physics C: Mechanics Gap)

AP Physics C covers projectile motion with calculus but does not specifically require memorisation of the trajectory equation y = gx²/(2u²) for horizontal projection or the ability to identify the coefficient of x² to extract initial speed. NEET tests: 'The trajectory of a horizontal projectile is y = 3x². Find the horizontal speed.' Requiring students to match to y = gx²/(2u²) → u = √(g/6).

y = gx²/(2u²) is the trajectory for horizontal launch — matching coefficients gives u
AP Physics C tests numerical projectile problems but not coefficient-matching with trajectory equations
NEET uses trajectory equations to set u and H conditions — a 2-step coefficient identification

NEET-Style Practice Questions — Horizontal Projectile Motion

4 Questions

1A ball is thrown horizontally from a 78.4 m high tower with speed 10 m/s. How far from the base of the tower does the ball strike the ground? (g = 9.8 m/s²)Horizontal Range in Horizontal Projection

20 m

30 m

40 m

50 m

T = √(2H/g) = √(2×78.4/9.8) = √(16) = 4 s. R = u × T = 10 × 4 = 40 m. The ball strikes 40 m from the base. Check: v_y = gT = 9.8×4 = 39.2 m/s. Speed at impact = √(100 + 1536.64) = √1636.64 ≈ 40.45 m/s.

2Three particles are released from the same height: A is dropped, B is thrown horizontally at 10 m/s, C is thrown horizontally at 20 m/s. Which statement is correct?Time of Flight in Horizontal Projection

A reaches the ground first because it has no horizontal velocity.

C reaches first because it has the highest speed.

All three particles reach the ground at the same time.

B and C reach at the same time, but A is slower.

The time of flight T = √(2H/g) depends only on the height H and gravitational acceleration g — it is completely independent of horizontal velocity. All three particles have zero initial vertical velocity (horizontal projection means θ = 0°, so u_y = 0 for B and C, and u = 0 for A). Therefore all three fall with the same vertical acceleration g and reach the ground simultaneously.

3A body is thrown horizontally from height H. After time T, it makes angle 30° with the horizontal. What is the horizontal velocity?Horizontal Range in Horizontal Projection

u = gT/√3

u = gT√3

u = gT/2

u = 2gT

At time T, vertical velocity v_y = gT. The angle with horizontal is 30°. tan(30°) = v_y/v_x = gT/u → 1/√3 = gT/u → u = gT√3. Alternatively: angle below horizontal = 30° means tan(30°) gives v_y/u. Since tan30° = 1/√3, u = v_y × √3 = gT√3.

4A ball is launched horizontally from a cliff and strikes the ground after 3 seconds with a speed of 50 m/s. What is the horizontal speed of the ball and the height of the cliff? (g = 10 m/s²)Trajectory of Horizontal Projectile

u = 30 m/s, H = 40 m

u = 40 m/s, H = 45 m

u = 50 m/s, H = 45 m

u = 30 m/s, H = 45 m

T = 3 s. Vertical velocity: v_y = gT = 30 m/s. Speed at impact: v = √(u² + v_y²) = 50 → u² + 900 = 2500 → u² = 1600 → u = 40 m/s. Height: H = (1/2)gT² = (1/2)(10)(9) = 45 m. Option B: u = 40 m/s, H = 45 m.

Practice Problems — Horizontal Projectile Motion

Click "Reveal Answer" after attempting

1From a height of 20 m, a ball is projected horizontally at 10 m/s. Find: (a) time of flight, (b) horizontal range. (g = 10 m/s²)

T=2s, R=20m

T=2s, R=10m

T=4s, R=20m

T=2s, R=40m

👁 Reveal Answer

Correct: T=2s, R=20m. T = √(2×20/10) = √4 = 2 s. R = 10 × 2 = 20 m.

2The trajectory of a horizontal projectile is y = 5x² (y in metres, x in metres). What is the horizontal speed if g = 10 m/s²?

1 m/s

√2 m/s

2 m/s

1/√2 m/s

👁 Reveal Answer

Correct: 1 m/s. From y = gx²/(2u²) = 5x²: g/(2u²) = 5 → u² = g/10 = 10/10 = 1 → u = 1 m/s.

3A stone is projected horizontally and after 2 s its velocity makes 45° with the horizontal. Find the initial horizontal velocity. (g = 10 m/s²)

10 m/s

20 m/s

5 m/s

40 m/s

👁 Reveal Answer

Correct: 20 m/s. After 2 s, v_y = gT = 20 m/s. Angle = 45° → tan45° = v_y/v_x = 1 → v_x = v_y = 20 m/s. Since horizontal velocity is constant: u = 20 m/s.

4Two stones are thrown horizontally from the top of a cliff: stone A at 5 m/s and stone B at 10 m/s. If stone A lands at a horizontal distance of 30 m, where does stone B land?

30 m

45 m

60 m

90 m

👁 Reveal Answer

Correct: 60 m. Time of flight T = R_A/u_A = 30/5 = 6 s (same for both since height is same). Range of B = u_B × T = 10 × 6 = 60 m. This confirms: for same height, R ∝ u.

Physics — Motion In Two Dimension Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Trajectory of Horizontal Projectile: Derive trajectory y = gx²/(2u²) by eliminating t from x = ut and y = (1/2)gt². Identify it as a parabola through the origin. Given y = 2x², find u when g = 10 m/s².

Time of Flight in Horizontal Projection: Write T = √(2H/g). Confirm T does not depend on horizontal speed u. Verify: from H = 20 m, T = 2 s for u = 5, 10, or 20 m/s. State the three-particle equal-time result and its explanation.

Horizontal Range in Horizontal Projection: Write R = u√(2H/g). Compute range and impact speed for: u = 15 m/s, H = 45 m. Find the angle velocity makes with horizontal at impact. Verify speed using energy conservation.

Impact Velocity — Magnitude and Direction: Write v_y = gT = √(2gH) and v = √(u² + 2gH). For a given u and H, find both the speed and angle at impact. Confirm this speed equals energy-conservation result √(u² + 2gH).

Equal-Time Three-Particle Conceptual Scenario: Write the assertion-reason pair: 'Particles A (dropped), B (horizontal at 10 m/s), C (horizontal at 20 m/s) from same height reach ground simultaneously.' Reason: time depends only on H. Both statements TRUE, reason is correct explanation.

FAQ — Horizontal Projectile Motion

Notes · Downloads · Revision · Important Questions

Does a ball thrown horizontally at higher speed take longer to reach the ground?

No. The time of flight T = √(2H/g) depends only on the height H from which it is thrown. Horizontal speed has no effect on the time to fall. A faster ball covers more horizontal distance in the same time — that is, it has greater range.

What is the shape of the trajectory of a horizontal projectile?

A parabola through the point of projection, opening downward. The equation is y = gx²/(2u²) where x is the horizontal distance and y is the vertical drop. The particle moves along this curve from (0, 0) downward and to the right.

How is the speed at impact calculated for horizontal projectile motion?

v_x = u (unchanged throughout). v_y = gT = g√(2H/g) = √(2gH). Speed at impact = √(v_x² + v_y²) = √(u² + 2gH). This equals the result from energy conservation: (1/2)mv² = (1/2)mu² + mgH → v = √(u² + 2gH). ✓

How do I find the angle of velocity with the horizontal at impact?

tan(α) = v_y/v_x = √(2gH)/u. This angle increases from 0° (at launch) to some angle α at impact. The angle depends on both u and H. For a steeper cliff (large H) or lower u, the angle at impact will be larger.

If a ball thrown horizontally and a ball dropped from the same height — which hits the ground first?

Both hit the ground at exactly the same time — T = √(2H/g) for both. The horizontal throw travels farther horizontally but takes the same time to fall the same vertical distance. This illustrates the principle of physical independence of horizontal and vertical motions.

How is horizontal projectile motion a special case of oblique projection?

In oblique projection: T = 2u sinθ/g. For θ = 0°: T = 0 (ball projected from ground). But for horizontal projection from a HEIGHT H, we use the free-fall formula T = √(2H/g). The oblique formula T = 2u sinθ/g applies when the ball returns to the same level as launch. These are two different scenarios — don't confuse them.

Can the trajectory equation be used to find the height of the launch point?

Not directly — the trajectory y = gx²/(2u²) gives vertical drop from the launch point, not absolute height. If x = R (at the ground), then y = H (height of cliff). Substituting: H = gR²/(2u²). Given R and u, this yields H. Or given the trajectory equation coefficients, identify g/(2u²) to find u.

What is the net velocity of a horizontal projectile at any time t?

v_x = u (constant). v_y = gt (increasing downward). Net speed: v(t) = √(u² + g²t²). The angle with horizontal: α(t) = arctan(gt/u). At t = 0: speed = u, angle = 0°. At t = T: speed = √(u² + 2gH), angle = arctan(√(2gH)/u).

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Horizontal Projectile Motion

NEET > Physics > Kinematics > Motion In Two Dimension > Horizontal Projectile Motion

Unit Progress

NEET Physics — Motion In Two Dimension

Horizontal Projectile Motion – Complete Notes, Revision, Important Questions & Downloads

⬇ Download Notes PDF View Important Questions →

7 Subtopicsθ = 0° CaseT = √(2H/g)

Expected QuestionsQ

0–1

Horizontal projection appears in NEET as the cliff-drop scenario or as a component of multi-step energy/2D questions approximately once every 2 years.

Time Required⏱

1.5 hrs

Difficulty⚡

Easy–Medium

NRI USA Curriculum GapUS

Low

7Subtopics

8+Practice Questions

4Free Downloads

1.5 hrsPrep Time

⬇ Get Free Downloads

NEET Weightage — Horizontal Projectile Motion

Motion In Two Dimension (Chapter 3)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
2022	1	1 Q	4
2021	0	0 Q	0
2020	1	1 Q	4
2019	0	0 Q	0
6-Year Total (2019–2024)	2–4		8–16

📊

0.5

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy–Medium

Difficulty

How to Prepare Horizontal Projection for NEET

Study Materials — Horizontal Projectile Motion

PDF · Cheat Sheet · MCQ Set · PYQ

📘

Full Notes

7 subtopics6 pagesDiagram-heavy

Download Notes

📗

Formula Sheet

y = gx²/(2u²); T = √(2H/g); R = u√(2H/g); v_y = √(2gH); v = √(u² + 2gH); tan(α) = √(2gH)/u. Table: u, H → T, R, v_impact.

6 formulas1 pageTable format

Download Sheet

📙

MCQ Practice

15 questions: find T given H, find R given u and H, find impact velocity, identify trajectory shape, equal-time scenario, and compare ranges from same height at different horizontal speeds.

15 MCQsGraded difficultySolved

Download MCQs

📒

PYQ

Year-tagged NEET previous-year horizontal projection questions with full solutions — cliff problems, equal-time assertions, and impact velocity calculations.

8+ year-tagged Qs2015–2024Step-by-step solutions

Download PYQs

Subtopics in Horizontal Projectile Motion

2-Column Table

Column A	Column B
Trajectory of Horizontal Projectile↗	Time of Flight in Horizontal Projection↗
Horizontal Range in Horizontal Projection↗	Units : second↗
Uniform Circular Motion↗	Angular displacement↗
Angular velocity↗

Rapid Revision — Horizontal Projectile Motion

Concept → Trap → Example

1) Trajectory of Horizontal Projectile

Core

Initial velocity: u (horizontal only, θ = 0°). Horizontal: x = ut. Vertical: y = (1/2)gt². Eliminate t: y = gx²/(2u²). This is a parabola through the origin, opening downward.

The trajectory y = gx²/(2u²) shows: y ∝ x². On the x-y plane (x = horizontal distance, y = vertical drop), the path curves downward.
For larger u: same x corresponds to smaller y (the projectile is flatter). The parabola is wider for higher initial speed.
Important: the vertex of the parabola is at the launch point (0,0). The path never goes above the launch height.

2) Time of Flight in Horizontal Projection

High Yield

Derivation: vertical motion from height H with u_y = 0 and a = g: H = (1/2)gT² → T = √(2H/g).
The three-particle result: A (dropped), B (horizontal 10 m/s), C (horizontal 20 m/s) from height H = 20 m all land at T = √(2×20/10) = 2 s simultaneously.
Trap: 'A ball projected horizontally at higher speed takes longer to fall' — FALSE. T = √(2H/g) is independent of u.

Example (NEET-style)From height H = 45 m: T = √(2×45/10) = √9 = 3 s. Any horizontal speed gives the same T = 3 s. Vertical velocity at ground: v_y = gT = 10×3 = 30 m/s downward.

3) Horizontal Range in Horizontal Projection

Core

Range R = u × T = u × √(2H/g) = u√(2H/g). Speed at impact: v = √(u² + v_y²) = √(u² + 2gH). Angle with horizontal: tan(α) = v_y/v_x = √(2gH)/u.

Range R ∝ u for fixed H: doubling horizontal speed doubles horizontal range.
Range R ∝ √H for fixed u: to double the range from same u, height must be quadrupled.
Speed at impact = √(u² + 2gH) — this equals the speed obtained by energy conservation, confirming that mechanical energy is conserved (gravity is conservative).

US Curriculum Gaps — Horizontal Projectile Motion

Topics in this section are tested in NEET but less emphasised in standard US physics courses.

Three-Particle Equal-Time Scenario (AP Physics 1 Gap)

AP Physics 1 covers the concept but does not test it as assertion-reason
NEET assertion: 'Three particles from same height take same time to land.' Reason: 'Time depends only on vertical free-fall.' Both statements true, reason is correct explanation.
This appears in NEET about once every 3 years — a reliable 4-mark question for students who drill it.

Trajectory Equation as a Parabola Through Origin (AP Physics C: Mechanics Gap)

y = gx²/(2u²) is the trajectory for horizontal launch — matching coefficients gives u
AP Physics C tests numerical projectile problems but not coefficient-matching with trajectory equations
NEET uses trajectory equations to set u and H conditions — a 2-step coefficient identification