« Projectile Motion: From absolute zero to exam hero »

Projectile Motion

From absolute zero to exam hero — the most visual, intuitive, and exam-ready guide to understanding parabolic motion.

Chapter 4 · NCERT
NEET Ready
JEE Concepts
Interactive Simulator

Exam Blueprint

Know exactly what CBSE and NEET expect. Here’s every subtopic mapped to its exam importance.

The Big Ideas

Understand these six ideas and you can solve any projectile problem. No memorisation — just intuition.

🎯

Two Independent Motions

Projectile motion = horizontal (constant speed) + vertical (free fall). They happen simultaneously but independently. Neither knows the other exists.

📐

Resolving Velocity

Any velocity at angle θ splits into Vx = u cosθ (horizontal) and Vy = u sinθ (vertical). This one skill unlocks every formula.

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At the Highest Point

Vy = 0 at the peak — ball momentarily stops moving up. But Vx = u cosθ remains. Speed at top = u cosθ. Never zero (for oblique projection).

Perfect Symmetry

The path is a perfect parabola. Time up = Time down. Speed at any height going up = Speed at same height going down. The first and second halves are mirror images.

🔗

Time is the Bridge

Time links horizontal and vertical motion. Find time first using vertical equations, then use it in the horizontal equation. This is the solving strategy.

♟️

Horizontal: No Force

ax = 0 always. No force acts horizontally (we ignore air resistance). So horizontal velocity never changes. This is Newton’s First Law in action.

Every Formula Explained

Not just formulas — derivation logic and memory tricks included. Understand them so you never forget them.

Time of Flight (Oblique)
T = 2u sinθ / g
Total time the projectile stays in the air
🧠 At top: Vy = 0 → 0 = usinθ − gt₁ → t₁ = usinθ/g. Total T = 2t₁
Maximum Height
H = u² sin²θ / 2g
Highest point above launch level
🧠 Use v² = u² − 2gH. At top v = 0: H = (usinθ)² / 2g
Horizontal Range
R = u² sin2θ / g
Horizontal distance from launch to landing
🧠 R = Vx × T = ucosθ × (2usinθ/g) = u²(2sinθcosθ)/g = u²sin2θ/g
Maximum Range
R_max = u² / g
Maximum possible range (at θ = 45°)
🧠 sin2θ is max = 1 when 2θ = 90° → θ = 45°. Then R = u²×1/g = u²/g
Horizontal Projection
T = √(2h/g), R = u√(2h/g)
Ball thrown horizontally from height h
🧠 Vertical: h = ½gT² → T = √(2h/g). Then R = horizontal speed × time
Equation of Trajectory
y = x tanθ − gx²/2u²cos²θ
Gives y at any horizontal distance x
🧠 Eliminate t: x = ucosθ·t → t = x/ucosθ. Substitute into y equation

♻️
Complementary Angles → Same Range
NEET Gift

If θ and (90°−θ) are two projection angles with the same initial speed, they always give the same range. Example: 30° & 60°, 20° & 70°, 25° & 65°.

Because: sin2(90°−θ) = sin(180°−2θ) = sin2θ → same R formula value

Bonus: For a complementary pair, H₁ × H₂ = R²/16 (JEE shortcut!)

Feel the Physics

Adjust speed and angle. Watch the ball fly. See every formula update in real time.

Initial Speed u (m/s)

20 m/s
Launch Angle θ

45°
Time of flight
2.83 s
Max height
10.0 m
Range
40.0 m
Vx (constant)
14.1 m/s

Try θ = 30° and θ = 60° with same speed → observe same range! · g = 10 m/s²

High-Scoring Scenarios

These specific setups appear repeatedly in NEET and CBSE. Recognise the pattern → apply the formula.

Horizontal Projection from Height

Ball thrown horizontally (θ = 0°) from a cliff of height h. No initial vertical velocity.

T = √(2h/g)
R = u · √(2h/g)
Vy at landing = √(2gh)
tan α = Vy/Vx = √(2gh)/u

Complementary Angles — Same Range

Two launch angles that sum to 90° always give identical range with same initial speed.

R(θ) = R(90°−θ)
H₁/H₂ = tan²θ₁ / tan²θ₂
H₁ · H₂ = R²/16
T₁/T₂ = sinθ₁/sinθ₂

Velocity at Any Point

At any time t, the velocity components and resultant speed can be found precisely.

Vx = u cosθ (always)
Vy = u sinθ − gt
Speed = √(Vx² + Vy²)
At top: speed = u cosθ

Don’t Fall for These

Students lose marks on these every year. Read each one once and never repeat them.

✗ Wrong

Using g = 10 m/s² in the horizontal direction. Writing ax = g or applying gravity to Vx.

✓ Correct

ax = 0 always. Gravity (g) only acts vertically downward. Horizontal velocity never changes because there’s no horizontal force.

✗ Wrong

Thinking velocity is zero at the highest point. Writing v = 0 at peak.

✓ Correct

Only Vy = 0 at the peak. Horizontal velocity Vx = u cosθ remains. Speed at top = u cosθ (never zero for oblique projection).

✗ Wrong

Sign confusion: Using +g when taking upward as positive, causing wrong answers in SUVAT equations.

✓ Correct

Define axis first. If upward = positive → g = −10 m/s². If downward = positive → g = +10 m/s². Be consistent throughout the entire problem.

✗ Wrong

For horizontal projection, using T = 2u sinθ/g. This gives T = 0 since sinθ = sin0° = 0.

✓ Correct

For horizontal projection: Use h = ½gT² → T = √(2h/g). The standard formula doesn’t apply when the initial vertical velocity is zero.

✗ Wrong

Confusing sinθ and cosθ when the angle is measured from the vertical instead of horizontal.

✓ Correct

Always check: angle from horizontal or vertical? « Projected from ground at angle θ » = from horizontal → Vx = ucosθ, Vy = usinθ. If from vertical, swap them.

Level Up Your Skills

Three difficulty levels. Click « Show Answer » only after attempting. The attempt matters more than the answer.

🟢 Level 1 — Basics
🟡 Level 2 — NCERT
🔴 Level 3 — NEET

Last Day Saver

Read this the morning of your exam. Everything you need in one place.

ALWAYS TRUE
ax = 0
No horizontal force
ALWAYS TRUE
ay = −g
Downward always
ALWAYS TRUE
Vx = u cosθ
Never changes

Oblique Projection

• T = 2u sinθ / g
• H = u² sin²θ / 2g
• R = u² sin2θ / g
• R_max = u²/g at θ = 45°
• Speed at top = u cosθ
• Speed at landing = launch speed

Horizontal Projection

• T = √(2h/g)
• R = u√(2h/g)
• Vy at ground = √(2gh)
• Speed at ground = √(u²+2gh)
• Landing angle: tanα = √(2gh)/u

Key Results

• Complementary angles → same R
• 30° & 60°: same range!
• At θ=45°: H = R/4
• H₁·H₂ = R²/16 (comp. pair)
• Trajectory: y = x tanθ(1−x/R)
• Radius at top = u²cos²θ/g

SUVAT for Vertical

• Vy = u sinθ − gt
• y = u sinθ·t − ½gt²
• Vy² = u²sin²θ − 2gy
• At top: Vy = 0
• Time up = Time down = T/2

Win on Exam Day

What to focus on, what patterns to expect, and how to save time in NEET/CBSE 2026.

CBSE Boards

• Derivation of T, H, R from SUVAT is frequently asked (2–3 marks each). Write all steps.
• NCERT Examples 4.7 and 4.8 are directly asked — solve them from the book.
• Horizontal projection from a height — very common in Section C (3 marks).
• Equation of trajectory derivation — a favourite long-answer question.
• Draw a neat diagram with labels in every answer — you lose 0.5 marks without it.

NEET 2026

• Expect 1–2 MCQs. Most likely: range, horizontal projection, or complementary angles.
• Complementary angles: tick same range in 5 seconds without calculating.
• Max range: answer is always u²/g at θ = 45°. Direct.
• Speed at top: immediately write u cosθ — 10 second question.
• If two numbers are given as angles summing to 90° — ranges are equal. Done.

JEE Concepts

• Equation of trajectory in parametric form — must derive smoothly.
• Radius of curvature at highest point = u²cos²θ/g.
• Projectile on inclined plane — study this separately (not in NCERT).
• Variable acceleration projectile — rare but appears in advanced papers.
• H₁·H₂ = R²/16 for complementary pairs — JEE loves this result.

Time-Saving Tricks

• « Angle for max range? » → Always θ = 45°. Zero calculations.
• « Complementary angles? » → Same R. Mark immediately.
• « Speed at top? » → u cosθ. Done in 5 seconds.
• Use g = 10 (not 9.8) unless the question specifies — much cleaner numbers.
• In MCQs: check units and eliminate options before solving fully.

Physics in Action

🏏

Cricket

A batsman hits at 45° for maximum distance — exactly as the range formula predicts. Bowlers use lower angles for faster deliveries reaching the batsman sooner.

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Football

A free-kick curling into the top corner follows a parabolic arc. Goalkeepers mentally estimate whether a shot clears the wall using projectile intuition.

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Rocketry & Satellites

Throw horizontally at 8 km/s — the ball falls but Earth curves away at the same rate. That’s an orbit. Projectile motion at a cosmic scale!

⛲

Water Fountains

Every water stream from a fountain is a projectile. Engineers calculate the angle and speed of water jets to achieve the exact parabolic arc desired.