Projectile Motion Master Hub
Master projectile motion from first principles to exam-level problem solving.
Learn the physics, derive the essential equations, recognise special cases,
practise numerical problems and revise the complete topic from one place.
🎯 Concepts
📐 Derivations
🧮 Numericals
📝 NEET Practice
⚡ Quick Revision
🚀 JEE Concepts
Jump to Formula Sheet
Practise NEET Questions
What Is Projectile Motion?
Projectile motion is the motion of an object projected into the air that
subsequently moves under gravity, when air resistance is neglected.
The motion can be understood as two simultaneous and independent motions:
uniform horizontal motion and vertically accelerated motion under gravity.
The central idea:
Split the initial velocity into horizontal and vertical components.
Solve the horizontal and vertical motions separately, then connect them
through time.
Projectile Motion Learning Path
① Understand the motion
→
② Resolve velocity
→
③ Derive equations
→
④ Recognise patterns
→
⑤ Solve numericals
→
⑥ Attempt NEET/JEE practice
1. Core Concepts
Two Independent Motions
Horizontal motion has constant velocity when air resistance is
neglected. Vertical motion is accelerated downward by gravity.
The two motions occur simultaneously but can be analysed separately.
Parabolic Trajectory
For ideal projectile motion near the Earth’s surface with constant
gravitational acceleration, the trajectory is a parabola.
Time Connects Both Motions
Time is the bridge between horizontal and vertical motion.
A reliable problem-solving method is to determine time from the
vertical motion and then use it in the horizontal motion.
At Maximum Height
The vertical component of velocity becomes zero at the highest point.
The horizontal component remains unchanged.
Symmetry
For launch and landing at the same level, the upward and downward
parts of the trajectory are symmetric. Time up equals time down.
Choose the Axis First
Define the positive direction before applying kinematic equations.
This prevents sign errors involving gravitational acceleration.
2. Resolving the Initial Velocity
Suppose a projectile is launched with speed u at an angle
θ above the horizontal.
Vx = u cos θ
Horizontal component — remains constant.
Vy = u sin θ
Initial vertical component.
Exam habit:
Always draw the velocity triangle before starting a difficult projectile
problem. It reduces sin/cos mistakes and makes the direction of each
component obvious.
3. Equations of Projectile Motion
Horizontal Motion
Vx = u cos θ
x = u cos θ · t
Horizontal acceleration is zero when air resistance is neglected.
Vertical Motion
Vy = u sin θ − gt
y = u sin θ · t − ½gt²
Vy² = u²sin²θ − 2gy
4. Complete Projectile Motion Formula Sheet
| Quantity | Formula | Key condition |
|---|---|---|
| Horizontal velocity | Vx = u cos θ | Constant |
| Initial vertical velocity | Vy = u sin θ | At launch |
| Time of flight | T = 2u sin θ / g | Launch and landing at same level |
| Time to maximum height | T/2 = u sin θ / g | Same-level oblique projection |
| Maximum height | H = u²sin²θ / 2g | Measured above launch level |
| Horizontal range | R = u²sin 2θ / g | Same launch and landing level |
| Maximum range | Rmax = u²/g | θ = 45° |
| Trajectory | y = x tan θ − gx²/(2u²cos²θ) | Ideal projectile |
| Horizontal projection time | T = √(2h/g) | Initial vertical velocity = 0 |
| Horizontal projection range | R = u√(2h/g) | Thrown horizontally from height h |
5. Special Cases & High-Yield Results
🏆 Maximum Range
Since sin 2θ is maximum when 2θ = 90°:
θ = 45°
Rmax = u²/g
♻️ Complementary Angles
Two angles θ and (90° − θ), for the same initial speed and
same-level launch and landing, produce the same range.
R(θ) = R(90° − θ)
Example: 30° and 60° produce the same range.
📌 Height = Range
Setting maximum height equal to range gives:
tan θ = 4
θ = tan⁻¹(4) ≈ 76°
🪂 Horizontal Projection
When a projectile is launched horizontally from height h,
its initial vertical velocity is zero.
T = √(2h/g)
R = u√(2h/g)
🔝 Speed at the Top
The vertical velocity becomes zero, but the horizontal velocity
remains.
Speed at top = u cos θ
📐 Complementary-Height Shortcut
For complementary projection angles with the same initial speed:
H₁ × H₂ = R² / 16
This is particularly useful for competitive-exam problems.
6. Worked Problem Types from the eduPhysics Projectile Collection
The older Projectile Motion articles contain several useful problem patterns.
Instead of maintaining them as overlapping theory articles, the Master Hub
now acts as the central learning map and sends students to the focused
worked-problem resources.
🌙 Earth vs Moon Range
A projectile launched with the same speed and angle has a different
range when the gravitational acceleration changes. The existing
eduPhysics example compares Earth with the Moon.
Key idea:
For the same u and θ, range is inversely proportional to g.
The existing example uses lunar gravity as approximately one-sixth
of Earth’s gravity.
View Earth–Moon Example
🏟️ Projectile Under a Ceiling
A projectile is launched inside a hall with a height restriction.
The problem requires the maximum allowable height first and then
the corresponding horizontal distance.
Key skill:
Combine the maximum-height equation with the range equation.
View Ceiling Problem
🧭 Perpendicular Velocity
A particle is projected at angle θ. Find the time at which its
instantaneous velocity becomes perpendicular to its initial
velocity.
Key technique:
Use the dot product of the initial and instantaneous velocity
vectors.
View Vector Problem
📝 NEET Projectile Practice
Use the dedicated NEET Projectile Motion Practice resource for
objective questions involving range, maximum height, time of flight,
trajectory and related exam patterns.
7. Practice, Quiz & Numerical Resources
NEET Projectile Motion Practice
Dedicated NEET-style projectile-motion questions and explanations.
Projectile Numericals
Use the numerical-practice pathway for calculation-based questions.
Projectile Quiz
Test conceptual understanding after completing the formula and
special-case sections.
8. Common Projectile Motion Mistakes
❌ Mistakes to Avoid
- Assuming horizontal acceleration is g.
- Thinking total velocity becomes zero at maximum height.
- Using the same time-of-flight formula for horizontal projection.
- Confusing sin θ and cos θ.
- Changing the sign convention midway through a problem.
- Using the same-level range formula when launch and landing heights differ.
✅ Better Problem-Solving Method
- Draw the projectile and define the axes.
- Resolve the initial velocity.
- Write horizontal and vertical equations separately.
- Identify the condition given in the question.
- Use time as the connection between x and y motion.
- Check units and limiting cases before finalising the answer.
9. Projectile Motion — Last-Day Revision
Always Remember
- ax = 0
- ay = −g when upward is positive
- Vx = u cos θ
- Vy = u sin θ − gt
- At the top, Vy = 0
High-Yield Results
- Maximum range → θ = 45°
- Complementary angles → same range
- Speed at top → u cos θ
- At 45°, H = R/4
- Time up = time down for same-level projection
Projectile Motion = Horizontal Motion + Vertical Motion
Master the two components and the topic becomes much easier to solve.
10. CBSE • NEET • JEE Exam Focus
📘 CBSE
- Derive time of flight, maximum height and range.
- Practise horizontal projection from a height.
- Understand the trajectory equation.
- Draw labelled diagrams in derivations and numerical solutions.
🧪 NEET
- Memorise the core formula relationships through understanding.
- Recognise complementary-angle questions immediately.
- Practise maximum range and horizontal-projection problems.
- Build speed with dedicated MCQ practice.
🚀 JEE
- Master trajectory derivations.
- Use complementary-angle relationships efficiently.
- Study vector-based projectile questions.
- Extend the basic model to advanced projectile configurations.
11. Related eduPhysics Resources
📚 Class 11 Motion in a Plane
Projectile motion is part of the broader Motion in a Plane framework.
Review vector components and two-dimensional kinematics before
attempting advanced projectile problems.
🎯 Projectile Motion Practice
Move from concept learning to exam practice with the dedicated
Projectile Motion question resource.
12. Projectile Motion FAQ
What is projectile motion?
Projectile motion is the motion of an object projected into the air
that subsequently moves under gravity when air resistance is neglected.
What is the trajectory of a projectile?
For ideal projectile motion near the Earth’s surface with constant
gravitational acceleration, the trajectory is a parabola.
Why is horizontal velocity constant?
In the ideal model, no horizontal force acts on the projectile.
Therefore horizontal acceleration is zero and horizontal velocity
remains constant.
Is velocity zero at the highest point?
No. Only the vertical component of velocity is zero. For an oblique
projectile, the horizontal component remains u cos θ.
At what angle is the range maximum?
For a projectile launched and landing at the same level with a fixed
initial speed and negligible air resistance, the maximum range occurs
at 45°.
Do complementary angles give the same range?
Yes. For the same initial speed and same launch and landing level,
complementary projection angles produce the same range.
What is the time of flight for horizontal projection?
For a projectile launched horizontally from height h,
T = √(2h/g).
How does gravity affect projectile range?
For the standard same-level formula, range is inversely proportional
to gravitational acceleration when initial speed and projection angle
are fixed.
What should I study after projectile motion basics?
Progress to numerical problem solving, horizontal projection,
complementary angles, trajectory problems, vector-based questions
and NEET/JEE practice.
solve worked examples → practise NEET questions → revise the formula sheet.
Start Projectile Practice
Review Motion in a Plane
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