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Projectile Motion Formulas for Class 11: Complete Derivation, Cheat Sheet & Solved Problems

NCERT Class 11 CBSE Board JEE Main & Advanced Chapter 4 — Motion in a Plane

Reviewed by a physics educator with 10+ years of experience in Class 11 mechanics and JEE preparation.

Contents

  1. —Formula Cheat Sheet
  2. 1What Is Projectile Motion?
  3. 2Two Independent Motions
  4. 3Time of Flight
  5. 4Maximum Height
  6. 5Horizontal Range
  7. 6Equation of Trajectory
  8. 7Velocity at Any Point
  9. 8Horizontal Projection
  10. 9Inclined Plane Case
  11. 10Solved: Board Level
  12. 11Solved: JEE Level
  13. 12Common Mistakes
  14. 13Practice Problems
  15. 14Summary Table
  16. 15FAQs

Quick-Reference Formula Cheat Sheet

Every formula on this page traces back to two basic kinematic equations — so once you understand the logic, you never need to memorise blindly. Bookmark this table for exam-eve revision.

Quantity Formula Key Variables
Time of Flight \( T = \dfrac{2u\sin\theta}{g} \) \(u\) = initial speed, \(\theta\) = launch angle
Maximum Height \( H = \dfrac{u^2\sin^2\!\theta}{2g} \) Peak vertical displacement
Horizontal Range \( R = \dfrac{u^2\sin 2\theta}{g} \) Horizontal distance at landing
Equation of Trajectory \( y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\!\theta} \) \(y\) as a function of \(x\) along the path
Speed at any time \(t\) \( v = \sqrt{v_x^2 + v_y^2} \) \(v_x = u\cos\theta\),   \(v_y = u\sin\theta - gt\)
Angle of velocity at time \(t\) \( \tan\alpha = \dfrac{v_y}{v_x} \) Direction of motion at any instant
📌 Exam Tip
For horizontal projection from height \(h\), formulas simplify because \(\theta = 0°\). Time of fall: \(t = \sqrt{2h/g}\)  ·  Horizontal range: \(R = u\sqrt{2h/g}\)

1. What Is Projectile Motion? Definition and Real-Life Examples

When you throw a cricket ball at an angle, it doesn't travel in a straight line. It curves upward, reaches a peak, then arcs back down. That curved path — and the physics behind it — is what we call projectile motion.

Here's the formal definition you'll find in NCERT:

A projectile is any object launched into space with an initial velocity, after which it moves under the influence of gravity alone — no engine, no thrust, no air resistance.

The key assumption in Class 11 is that air resistance is negligible. This simplification is what makes the mathematics clean and the trajectory perfectly parabolic.

Real-life examples students can visualise:

🏏A cricket ball hit at an angle — its arc from bat to boundary
🚿Water arcing out of a garden hose held at an angle
🏀A basketball released toward the hoop
🪨A stone rolled off the edge of a table (horizontal projection)
🏹A javelin thrown at a sports event

Once you understand the formulas, you can predict exactly where any of these objects will land.

2. The Core Idea: Two Independent Motions Happening Simultaneously

This is the single most important concept in this entire topic. If you understand this, the rest follows naturally.

When a ball is thrown at an angle, it doesn't have one complicated motion — it has two simple motions happening at the same time:

Horizontal Motion (x-direction)

  • No horizontal force acts on the object
  • Horizontal velocity stays constant
  • \(u_x = u\cos\theta\),   \(a_x = 0\)

Vertical Motion (y-direction)

  • Gravity pulls downward with \(g = 9.8\) m/s²
  • Vertical velocity changes every second
  • \(u_y = u\sin\theta\),   \(a_y = -g\)
Key Insight
These two motions are completely independent of each other. The ball doesn't "know" it's moving forward while it's falling downward. This independence is why we can split every projectile problem into two separate 1D problems.

The three kinematic equations used throughout this topic:

\( v = u + at \)
\( s = ut + \tfrac{1}{2}at^2 \)
\( v^2 = u^2 - 2as \)

You already know these from Chapter 3. Projectile motion simply applies them in two directions at once.

3. Time of Flight Formula: Derivation and Meaning

Time of flight (T) is the total time the projectile stays in the air — from launch to when it hits the ground at the same horizontal level.

Derivation

At the end of flight, the projectile returns to its original height. So the net vertical displacement is zero.

1
Apply \(s = ut + \frac{1}{2}at^2\) in the vertical direction with \(s = 0\):
\[ 0 = (u\sin\theta)\,T - \tfrac{1}{2}g T^2 \]
2
Factor out \(T\):
\[ 0 = T\!\left(u\sin\theta - \tfrac{1}{2}gT\right) \]
3
Since \(T \neq 0\), the bracket must be zero:
\[ u\sin\theta = \tfrac{1}{2}gT \]
4
Solve for \(T\):
\[ \boxed{T = \frac{2u\sin\theta}{g}} \]

What This Formula Tells You

⚠ Common Mistake
Many students write \(T = u\sin\theta / g\) — missing the factor of 2. The factor of 2 appears because the projectile takes equal time going up and coming down. Time to reach peak is \(T/2 = u\sin\theta/g\); total time is twice that.

4. Maximum Height Formula: How to Derive and Apply It

Maximum height (H) is the highest point the projectile reaches above its launch level.

Derivation

At the highest point, the vertical velocity becomes zero — the object momentarily stops moving upward before falling back down.

1
Apply \(v^2 = u^2 - 2as\) vertically with final vertical velocity \(= 0\):
\[ 0 = (u\sin\theta)^2 - 2gH \]
2
Rearrange:
\[ 2gH = u^2\sin^2\!\theta \]
3
Result:
\[ \boxed{H = \frac{u^2\sin^2\!\theta}{2g}} \]
Key Insight
Maximum height depends entirely on the vertical component of initial velocity. The horizontal component plays no role. At \(\theta = 90°\) (straight up), \(H = u^2/2g\) — the maximum possible height for a given speed.

5. Horizontal Range Formula: Derivation and the Angle That Maximises It

Horizontal range (R) is the total horizontal distance covered when the projectile lands at the same height it was launched from.

Derivation

1
Range = horizontal velocity × total time of flight:
\[ R = u_x \times T = (u\cos\theta)\times\frac{2u\sin\theta}{g} \]
2
Simplify using the identity \(2\sin\theta\cos\theta = \sin 2\theta\):
\[ \boxed{R = \frac{u^2\sin 2\theta}{g}} \]

The Angle for Maximum Range

Since \(\sin 2\theta\) reaches its maximum value of 1 when \(2\theta = 90°\), i.e., \(\theta = 45°\):

Maximum range (at θ = 45°)
\[ R_{\max} = \frac{u^2}{g} \]

Complementary Angles Give Equal Range

Since \(\sin 2(30°) = \sin 60°\) and \(\sin 2(60°) = \sin 120° = \sin 60°\), a ball thrown at 30° and another at 60° (same speed) land at exactly the same spot. This is a frequently tested result.

Launch Anglesin 2θRelative Range
15°\(\sin 30° = 0.5\)Half of maximum
30°\(\sin 60° = 0.866\)86.6% of maximum
45°\(\sin 90° = 1.0\)Maximum range
60°\(\sin 120° = 0.866\)Same as 30°
75°\(\sin 150° = 0.5\)Same as 15°

6. Equation of Trajectory: Why the Path Is Always a Parabola

The equation of trajectory gives the shape of the path — the \(y\)-coordinate at every \(x\)-coordinate along the flight.

Derivation

1
Horizontal position:   \(x = (u\cos\theta)\,t\),   so   \(t = \dfrac{x}{u\cos\theta}\)
2
Vertical position:   \(y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2\)
3
Substitute \(t = x/(u\cos\theta)\) into the vertical equation:
\[ y = (u\sin\theta)\cdot\frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2 \]
4
Simplify to the trajectory equation:
\[ \boxed{y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\!\theta}} \]

This is in the form \(y = ax + bx^2\) — the standard equation of a parabola.

Practical Use
Use this equation when a problem asks: "At what horizontal distance does the ball reach a height of 10 m?" — substitute \(y = 10\) and solve for \(x\).

7. Velocity at Any Point During Flight

At any time \(t\) during the flight:

Horizontal component (constant)

  • \(v_x = u\cos\theta\)

Vertical component (changes)

  • \(v_y = u\sin\theta - gt\)
Resultant speed
\[ v = \sqrt{v_x^2 + v_y^2} \]
Direction with horizontal
\[ \tan\alpha = \frac{v_y}{v_x} \]
📌 Remember
At the highest point, \(v_y = 0\), so speed is at its minimum and equals \(u\cos\theta\). The velocity there is purely horizontal — never zero (unless \(\theta = 90°\)).

8. Horizontal Projection from a Height

A very common special case: the object is launched horizontally (\(\theta = 0°\)) from a height \(h\). Think of a ball rolling off a table, or a stone pushed horizontally off a cliff.

Since \(\theta = 0°\):   \(u\sin\theta = 0\) (no initial vertical velocity) and \(u\cos\theta = u\) (all speed is horizontal).

1
Time to fall height \(h\) (vertical displacement = \(h\), initial vertical velocity = 0):
\[ h = \tfrac{1}{2}gt^2 \implies \boxed{t = \sqrt{\frac{2h}{g}}} \]
2
Horizontal range:
\[ \boxed{R = u\,t = u\sqrt{\frac{2h}{g}}} \]
3
Equation of trajectory (vertex at launch point):
\[ y = \frac{g\,x^2}{2u^2} \quad \text{(parabola opening downward)} \]
Surprising but True
A ball pushed fast off a table and a ball pushed slowly off the same table both hit the floor at the same time. The time of fall depends only on the height and \(g\) — not on horizontal speed. Only the landing distance differs.

9. Projectile on an Inclined Plane

This is a more advanced topic, tested frequently in JEE Main and JEE Advanced. When the surface is inclined at angle \(\beta\), we resolve motion along the incline and perpendicular to it instead of the standard horizontal/vertical axes.

Setup (projectile launched up the incline at angle α above the incline)

Along incline (opposing motion)

  • Retardation: \(g\sin\beta\)

Perpendicular to incline

  • Retardation: \(g\cos\beta\)
Time of flight (up the incline)
\[ T = \frac{2u\sin\alpha}{g\cos\beta} \]
Range along the incline
\[ R = \frac{2u^2\sin\alpha\,\cos(\alpha+\beta)}{g\cos^2\!\beta} \]
Angle for maximum range on incline
\[ \alpha = 45° - \frac{\beta}{2} \]
📌 Approach
For motion down the incline, the component of \(g\) along the slope now assists motion (not opposes). Replace \(g\sin\beta\) with \(-g\sin\beta\) in the along-incline equation and re-derive.

10. Solved Examples: NCERT and Board Level

P1 Time of flight and range Board Level
A ball is thrown at 30° with the horizontal at an initial speed of 20 m/s. Find (a) the time of flight and (b) the horizontal range. (\(g = 10\) m/s²)
(a) Time of flight: \[ T = \frac{2u\sin\theta}{g} = \frac{2\times 20\times\sin 30°}{10} = \frac{2\times 20\times 0.5}{10} = 2\text{ s} \]
(b) Horizontal range: \[ R = \frac{u^2\sin 2\theta}{g} = \frac{(20)^2\times\sin 60°}{10} = \frac{400\times 0.866}{10} \approx 34.6\text{ m} \]
Answers: T = 2 s, R ≈ 34.6 m
P2 Range equals maximum height — find the angle Board Level
Find the angle of projection for which the horizontal range equals the maximum height.
Set \(R = H\): \[ \frac{u^2\sin 2\theta}{g} = \frac{u^2\sin^2\!\theta}{2g} \]
Use \(\sin 2\theta = 2\sin\theta\cos\theta\): \[ 2\sin\theta\cos\theta = \frac{\sin^2\!\theta}{2} \implies 4\cos\theta = \sin\theta \implies \tan\theta = 4 \]
Answer: θ = arctan(4) ≈ 76°
P3 Stone thrown from a tower Board Level
A stone is thrown at 19.6 m/s at 30° above the horizontal from the top of a 490 m tower. Find (a) time in air and (b) horizontal distance from the base where it lands. (\(g = 9.8\) m/s²)
Initial vertical velocity (upward): \(u_y = 19.6\sin 30° = 9.8\) m/s
Vertical displacement when landing: \(s = -490\) m (downward from top)
Use \(s = u_y t - \tfrac{1}{2}gt^2\): \[ -490 = 9.8t - 4.9t^2 \implies t^2 - 2t - 100 = 0 \] \[ t = \frac{2 + \sqrt{4 + 400}}{2} = 1 + \sqrt{101} \approx 11.05\text{ s} \]
Horizontal range: \(R = u\cos 30° \times t = 19.6\times 0.866\times 11.05 \approx 187.5\text{ m}\)
Answers: t ≈ 11.05 s, R ≈ 187.5 m

11. Solved Examples: JEE Main and JEE Advanced Level

P4 Speed at greatest height vs. half maximum height JEE Classic
The speed of a projectile at its greatest height is \(\sqrt{2/5}\) times its speed at half the maximum height. Find the angle of projection.
At the highest point only horizontal speed remains:   \(v_H^2 = u^2\cos^2\!\theta \quad\cdots(1)\)
At half the maximum height \((h = H/2)\): \[ v_{H/2}^2 = u^2 - 2g\!\cdot\!\frac{H}{2} = u^2 - gH = u^2 - \frac{u^2\sin^2\!\theta}{2} \quad\cdots(2) \]
Apply the given condition \(v_H = \sqrt{2/5}\cdot v_{H/2}\), so \(v_H^2 = \tfrac{2}{5}\,v_{H/2}^2\): \[ u^2\cos^2\!\theta = \tfrac{2}{5}\!\left(u^2 - \tfrac{u^2\sin^2\!\theta}{2}\right) \] \[ 5\cos^2\!\theta = 2 - \sin^2\!\theta \implies 5(1-\sin^2\!\theta) = 2 - \sin^2\!\theta \] \[ 4\sin^2\!\theta = 3 \implies \sin\theta = \frac{\sqrt{3}}{2} \implies \theta = 60° \]
Answer: θ = 60°

Thought process: Whenever JEE gives a ratio of speeds at two heights, immediately write \(v^2 = u^2 - 2gh\) expressions for both points, then apply the given ratio. Don't separate components — that leads to unnecessary algebra.

P5 Two projectiles — are they ever at the same height? JEE Level
Two projectiles are launched simultaneously from the same point — one at 60° and one at 30° — both with the same initial speed \(u\). At what time \(t\) are they at the same height?
Height equations: \[ y_1 = u\sin 60°\cdot t - \tfrac{1}{2}gt^2, \quad y_2 = u\sin 30°\cdot t - \tfrac{1}{2}gt^2 \]
Set \(y_1 = y_2\): the \(\tfrac{1}{2}gt^2\) terms cancel: \[ u\sin 60°\cdot t = u\sin 30°\cdot t \] This holds only if \(t = 0\) or \(\sin 60° = \sin 30°\) — the latter is never true.
Answer: They are never at the same height after launch (except t = 0)

Lesson: Don't assume two projectiles from the same point are ever at equal height just because they end up at the same landing spot. Complementary angles give equal range, not equal height at any intermediate time.

12. High-Weightage Topics and Common Mistakes to Avoid

Most commonly tested in exams

  1. Finding range, time of flight, or max height given initial speed and angle
  2. The 45° maximum range result and complementary angle pairs
  3. Horizontal projection problems (\(\theta = 0\), launched from a height)
  4. Equation of trajectory — substituting a specific \(x\) or \(y\) value
  5. Speed and direction of velocity at a given point (especially at the peak)

Top 5 mistakes that cost marks

📌 Memory Tip: T → H → R → Trajectory
  • T = \(2u\sin\theta/g\) — from vertical displacement = 0
  • H = \(u^2\sin^2\!\theta/2g\) — from vertical velocity = 0 at peak
  • R = \(u^2\sin 2\theta/g\) — horizontal speed × T, then double-angle identity
  • Trajectory — eliminate time between \(x\) and \(y\) equations

13. Practice Problems: Test Yourself Before the Exam

Try each problem before revealing the hint. They cover the full difficulty range from board level to JEE Advanced.

Problem A Easy
A ball is projected at 10 m/s at 45° to the horizontal. Find the maximum height reached. (\(g = 10\) m/s²)
Show hint & answer
Use \(H = u^2\sin^2\!\theta / 2g\) with \(u = 10\) m/s, \(\theta = 45°\): \[ H = \frac{100 \times 0.5}{20} = \frac{50}{20} = 2.5\text{ m} \] Answer: 2.5 m
Problem B Easy
Show that the time of flight of a projectile is twice the time it takes to reach maximum height.
Show hint
Time to reach max height (vertical velocity = 0): \(t_{\text{peak}} = u\sin\theta/g\). Total time of flight: \(T = 2u\sin\theta/g = 2\,t_{\text{peak}}\).   ∎
Problem C Medium
A projectile is fired at 60° and travels a horizontal range of 100 m. What is the initial speed? (\(g = 10\) m/s²)
Show hint & answer
\[ R = \frac{u^2\sin 2\theta}{g} \implies 100 = \frac{u^2\sin 120°}{10} = \frac{u^2\times 0.866}{10} \] \[ u^2 = \frac{1000}{0.866} \approx 1155 \implies u \approx 34\text{ m/s} \] Answer: u ≈ 34 m/s
Problem D Medium
A ball is thrown horizontally at 20 m/s from a tower of height 80 m. Find (a) the time to land and (b) the horizontal distance from the base. (\(g = 10\) m/s²)
Show hint & answer
(a) \(t = \sqrt{2h/g} = \sqrt{2\times 80/10} = \sqrt{16} = 4\text{ s}\)
(b) \(R = u\times t = 20\times 4 = 80\text{ m}\)
Answers: (a) 4 s   (b) 80 m
Problem E Hard — JEE Advanced
A particle is projected at angle \(\theta\) such that its range equals twice its maximum height. Find \(\theta\).
Show hint & answer
Set \(R = 2H\): \[ \frac{u^2\sin 2\theta}{g} = 2\cdot\frac{u^2\sin^2\!\theta}{2g} = \frac{u^2\sin^2\!\theta}{g} \] \[ \sin 2\theta = \sin^2\!\theta \implies 2\sin\theta\cos\theta = \sin^2\!\theta \implies 2\cos\theta = \sin\theta \implies \tan\theta = 2 \] Answer: θ = arctan(2) ≈ 63.4°

14. Summary: All Projectile Motion Formulas at a Glance

Standard projectile (launched at angle θ from a horizontal surface)

QuantityFormula
Horizontal velocity (constant)\(v_x = u\cos\theta\)
Vertical velocity at time \(t\)\(v_y = u\sin\theta - gt\)
Horizontal displacement at \(t\)\(x = (u\cos\theta)\,t\)
Vertical displacement at \(t\)\(y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2\)
Time of flight\(T = 2u\sin\theta/g\)
Maximum height\(H = u^2\sin^2\!\theta\,/\,2g\)
Horizontal range\(R = u^2\sin 2\theta\,/\,g\)
Maximum range (θ = 45°)\(R_{\max} = u^2/g\)
Equation of trajectory\(y = x\tan\theta - gx^2/(2u^2\cos^2\!\theta)\)

Horizontal projection from height h (θ = 0°)

QuantityFormula
Time to land\(t = \sqrt{2h/g}\)
Horizontal range\(R = u\sqrt{2h/g}\)
Equation of trajectory\(y = gx^2/2u^2\)
Foundation
Every formula above can be derived from just two equations:   \(s = ut + \tfrac{1}{2}at^2\) and \(v^2 = u^2 - 2as\). Apply these in the \(x\) and \(y\) directions separately, and you can re-derive anything from scratch in an exam.

15. Frequently Asked Questions

Why is the trajectory of a projectile always a parabola?
Because horizontal motion is uniform (constant velocity) and vertical motion is uniformly accelerated (constant \(g\)). When you eliminate time from the two position equations, you get \(y\) as a quadratic function of \(x\) — exactly the definition of a parabola. Any combination of constant velocity in one direction and constant acceleration in a perpendicular direction will always produce a parabolic path.
Does air resistance affect the range and maximum height?
Yes, significantly. In the real world, air resistance acts opposite to the direction of motion and reduces both range and maximum height. It also makes the trajectory asymmetric — the descent is steeper than the ascent. For Class 11 and JEE, we assume air resistance is negligible, which gives the clean parabolic path. In real engineering and ballistics, air resistance requires numerical methods to model accurately.
Is projectile motion part of the NCERT Class 11 syllabus?
Yes. Projectile motion is covered in Chapter 4 (Motion in a Plane) of the NCERT Class 11 Physics textbook. CBSE board exams regularly include 3–5 mark questions on this topic. For JEE Main it appears as direct formula application; for JEE Advanced it often appears as multi-concept problems combining projectile motion with energy conservation, relative motion, or inclined planes.
What angle gives the maximum range on a horizontal surface?
45°. At this angle, \(\sin 2\theta = \sin 90° = 1\), which is its maximum possible value. Any angle above or below 45° gives a smaller range for the same initial speed. The maximum range is \(u^2/g\).
How is projectile motion different from circular motion?
In projectile motion, the path is parabolic and the speed changes over time (only the vertical component changes; horizontal stays constant). In circular motion, the path is a circle and the direction changes continuously. Both involve acceleration — in projectile motion, acceleration is constant and always directed downward (gravity); in uniform circular motion, acceleration is centripetal, always pointing toward the centre and therefore constantly changing direction.
At the highest point, is the acceleration of the projectile zero?
No — this is one of the most common misconceptions. At the highest point, the vertical velocity is zero, but gravity still acts on the object. The acceleration is still \(g = 9.8\) m/s² downward. Velocity being zero at an instant does not mean acceleration is zero. The object is decelerating on the way up and accelerating downward on the way down — gravity never switches off mid-flight.