The Projectile Motion Formula describes the trajectory, time of flight, maximum height, and horizontal range of any object launched into the air under gravity alone. Expressed as a set of kinematic equations, this formula is a core topic in NCERT Class 11 Physics (Chapter 4 — Motion in a Plane). It is equally vital for JEE Main, JEE Advanced, and NEET, where questions on projectile motion appear almost every year. This article covers every key expression, step-by-step derivations, three progressive solved examples, a complete formula sheet, CBSE exam tips for 2025-26, and common mistakes to avoid.
Key Projectile Motion Formulas at a Glance
Quick reference for the most important projectile motion expressions.
- Horizontal velocity: \( u_x = u\cos\theta \)
- Vertical velocity: \( u_y = u\sin\theta \)
- Time of flight: \( T = \dfrac{2u\sin\theta}{g} \)
- Maximum height: \( H = \dfrac{u^2\sin^2\theta}{2g} \)
- Horizontal range: \( R = \dfrac{u^2\sin 2\theta}{g} \)
- Equation of trajectory: \( y = x\tan\theta – \dfrac{gx^2}{2u^2\cos^2\theta} \)
- Velocity at any instant: \( v = \sqrt{v_x^2 + v_y^2} \)
What is Projectile Motion Formula?
The Projectile Motion Formula is a collection of kinematic equations that govern the two-dimensional motion of an object launched at an angle to the horizontal, moving freely under the influence of gravity alone. No other force — such as air resistance — acts on the projectile once it is in flight.
This concept is introduced in NCERT Class 11 Physics, Chapter 4 (Motion in a Plane). The key insight is that projectile motion can be resolved into two independent components: uniform horizontal motion (no acceleration) and uniformly accelerated vertical motion (acceleration = g downward).
Because these two components are independent, we apply separate equations of motion to each direction. The combination of these equations gives us the trajectory, which turns out to be a parabola. Understanding the Projectile Motion Formula is essential not only for CBSE board exams but also for solving mechanics problems in JEE Main and NEET, where this topic regularly contributes 1–2 questions per exam.
Projectile Motion Formula — Expression and Variables
Consider a projectile launched with initial speed \( u \) at an angle \( \theta \) above the horizontal. The standard expressions are:
Horizontal and Vertical Components of Initial Velocity:
\[ u_x = u\cos\theta, \quad u_y = u\sin\theta \]
Position at time t:
\[ x = u\cos\theta \cdot t, \quad y = u\sin\theta \cdot t – \frac{1}{2}gt^2 \]
Velocity at time t:
\[ v_x = u\cos\theta, \quad v_y = u\sin\theta – gt \]
Time of Flight (T):
\[ T = \frac{2u\sin\theta}{g} \]
Maximum Height (H):
\[ H = \frac{u^2\sin^2\theta}{2g} \]
Horizontal Range (R):
\[ R = \frac{u^2\sin 2\theta}{g} \]
Equation of Trajectory:
\[ y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( u \) | Initial speed of projection | m/s |
| \( \theta \) | Angle of projection with horizontal | degrees or radians |
| \( u_x \) | Horizontal component of initial velocity | m/s |
| \( u_y \) | Vertical component of initial velocity | m/s |
| \( g \) | Acceleration due to gravity | m/s² |
| \( t \) | Time elapsed | s |
| \( T \) | Total time of flight | s |
| \( H \) | Maximum height reached | m |
| \( R \) | Horizontal range | m |
| \( x, y \) | Horizontal and vertical displacement | m |
| \( v_x, v_y \) | Horizontal and vertical velocity components at time t | m/s |
Derivation of Projectile Motion Formulas
A projectile is launched from the origin with speed \( u \) at angle \( \theta \). We resolve motion into two directions.
Step 1 — Horizontal: No horizontal force acts, so \( a_x = 0 \). Using \( x = u_x t \), we get \( x = u\cos\theta \cdot t \).
Step 2 — Vertical: Gravity acts downward, so \( a_y = -g \). Using \( y = u_y t – \frac{1}{2}gt^2 \), we get \( y = u\sin\theta \cdot t – \frac{1}{2}gt^2 \).
Step 3 — Time of Flight: At landing, \( y = 0 \). Solving gives \( T = \frac{2u\sin\theta}{g} \).
Step 4 — Maximum Height: At the top, \( v_y = 0 \). Using \( v_y^2 = u_y^2 – 2gH \), we get \( H = \frac{u^2\sin^2\theta}{2g} \).
Step 5 — Range: Substituting \( T \) into the horizontal equation gives \( R = \frac{u^2\sin 2\theta}{g} \).
Step 6 — Trajectory: Eliminating \( t \) from the position equations yields the parabolic path \( y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta} \).
Complete Physics Formula Sheet — Projectile Motion
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Horizontal velocity (constant) | \( v_x = u\cos\theta \) | u = initial speed, θ = angle | m/s | Class 11, Ch 4 |
| Vertical velocity at time t | \( v_y = u\sin\theta – gt \) | g = 9.8 m/s², t = time | m/s | Class 11, Ch 4 |
| Horizontal displacement | \( x = u\cos\theta \cdot t \) | t = time elapsed | m | Class 11, Ch 4 |
| Vertical displacement | \( y = u\sin\theta \cdot t – \frac{1}{2}gt^2 \) | g = acceleration due to gravity | m | Class 11, Ch 4 |
| Time of flight | \( T = \frac{2u\sin\theta}{g} \) | u, θ, g | s | Class 11, Ch 4 |
| Maximum height | \( H = \frac{u^2\sin^2\theta}{2g} \) | u, θ, g | m | Class 11, Ch 4 |
| Horizontal range | \( R = \frac{u^2\sin 2\theta}{g} \) | u, θ, g | m | Class 11, Ch 4 |
| Maximum range (at θ = 45°) | \( R_{max} = \frac{u^2}{g} \) | u, g | m | Class 11, Ch 4 |
| Equation of trajectory | \( y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta} \) | x, y = coordinates; u, θ, g | m | Class 11, Ch 4 |
| Speed at any instant | \( v = \sqrt{v_x^2 + v_y^2} \) | v_x, v_y = velocity components | m/s | Class 11, Ch 4 |
| Angle of velocity with horizontal at time t | \( \phi = \tan^{-1}\!\left(\frac{v_y}{v_x}\right) \) | v_x, v_y | degrees | Class 11, Ch 4 |
| Height at horizontal distance x | \( y = x\tan\theta\left(1 – \frac{x}{R}\right) \) | x, R = range, θ | m | Class 11, Ch 4 |
Projectile Motion Formula — Solved Examples
Example 1 (Class 9–10 Level): Finding Range and Time of Flight
Problem: A ball is kicked with an initial speed of 20 m/s at an angle of 30° above the horizontal. Find (a) the time of flight and (b) the horizontal range. (Take g = 10 m/s².)
Given: u = 20 m/s, θ = 30°, g = 10 m/s²
Step 1: Find the vertical component of initial velocity.
\( u_y = u\sin\theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10 \) m/s
Step 2: Apply the time of flight formula.
\( T = \dfrac{2u_y}{g} = \dfrac{2 \times 10}{10} = 2 \) s
Step 3: Find the horizontal component of initial velocity.
\( u_x = u\cos\theta = 20 \times \cos 30^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 \) m/s
Step 4: Apply the range formula.
\( R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{400 \times \sin 60^\circ}{10} = \dfrac{400 \times 0.866}{10} = 34.64 \) m
Answer
Time of flight T = 2 s; Horizontal range R ≈ 34.64 m
Example 2 (Class 11–12 Level): Maximum Height and Velocity at Highest Point
Problem: A stone is projected with an initial velocity of 40 m/s at 60° to the horizontal. Calculate (a) the maximum height reached and (b) the speed of the stone at the highest point. (g = 10 m/s².)
Given: u = 40 m/s, θ = 60°, g = 10 m/s²
Step 1: Resolve initial velocity.
\( u_x = 40\cos 60^\circ = 40 \times 0.5 = 20 \) m/s
\( u_y = 40\sin 60^\circ = 40 \times \dfrac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.64 \) m/s
Step 2: Apply the maximum height formula.
\( H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{1600 \times \sin^2 60^\circ}{2 \times 10} = \dfrac{1600 \times 0.75}{20} = \dfrac{1200}{20} = 60 \) m
Step 3: Find the speed at the highest point.
At the maximum height, the vertical velocity component is zero. Only the horizontal component remains.
\( v_{top} = u_x = u\cos\theta = 20 \) m/s
Step 4: Verify using energy (optional cross-check).
\( v_{top}^2 = u^2 – 2gH = 1600 – 2 \times 10 \times 60 = 1600 – 1200 = 400 \)
\( v_{top} = 20 \) m/s ✓
Answer
Maximum height H = 60 m; Speed at highest point = 20 m/s
Example 3 (JEE/NEET Level): Trajectory and Complementary Angles
Problem: A projectile is fired with speed u = 30 m/s. (a) Show that two angles of projection θ and (90° − θ) give the same range. (b) If the range is 45 m and g = 10 m/s², find the two possible angles of projection. Also find the ratio of the maximum heights for these two angles.
Given: u = 30 m/s, R = 45 m, g = 10 m/s²
Step 1: Use the range formula for angle θ.
\( R_1 = \dfrac{u^2\sin 2\theta}{g} \)
Step 2: Use the range formula for angle (90° − θ).
\( R_2 = \dfrac{u^2\sin(180^\circ – 2\theta)}{g} = \dfrac{u^2\sin 2\theta}{g} = R_1 \)
This confirms that complementary angles yield the same range.
Step 3: Find θ using the given range.
\( 45 = \dfrac{900 \times \sin 2\theta}{10} \Rightarrow \sin 2\theta = \dfrac{45 \times 10}{900} = 0.5 \)
\( 2\theta = 30^\circ \Rightarrow \theta_1 = 15^\circ \) and \( \theta_2 = 90^\circ – 15^\circ = 75^\circ \)
Step 4: Find the ratio of maximum heights.
\( H_1 = \dfrac{u^2\sin^2 15^\circ}{2g} \) and \( H_2 = \dfrac{u^2\sin^2 75^\circ}{2g} \)
\( \dfrac{H_1}{H_2} = \dfrac{\sin^2 15^\circ}{\sin^2 75^\circ} = \tan^2 15^\circ \approx \dfrac{(0.2679)^2}{(0.9659)^2} \approx \dfrac{0.0718}{0.933} \approx 0.077 \approx \dfrac{1}{13} \)
More precisely, \( \dfrac{H_1}{H_2} = \tan^2\theta = \tan^2 15^\circ \). This is a standard JEE result: for complementary projection angles giving the same range, \( H_1 H_2 = \dfrac{R^2}{16} \) and \( H_1 + H_2 = \dfrac{R}{4}\tan\theta + \dfrac{R}{4}\cot\theta \).
Answer
The two projection angles are 15° and 75°. Both give a range of 45 m. The ratio H&sub1;/H&sub2; = tan²15° ≈ 1:13. Complementary angles always produce equal ranges — a key JEE/NEET result.
CBSE Exam Tips 2025-26
- Memorise all six core expressions. The time of flight, maximum height, range, and trajectory formulas are directly asked in 2–3 mark questions. Write them on a revision card.
- Always resolve velocity first. In every projectile problem, the first step must be splitting the initial velocity into \( u\cos\theta \) and \( u\sin\theta \). We recommend writing this step explicitly to earn method marks.
- State g clearly. CBSE papers sometimes specify g = 9.8 m/s² or g = 10 m/s². Use whichever value the question provides. Mixing them costs marks.
- Draw the trajectory diagram. A neat parabolic sketch with labelled H, R, and T earns 1 bonus mark in 5-mark problems. Our experts suggest spending 30 seconds on the diagram.
- Learn the complementary angle result. The fact that \( R(\theta) = R(90^\circ – \theta) \) is a frequently tested one-liner in CBSE MCQs and assertion-reason questions in 2025-26.
- Practice unit conversion. If initial speed is given in km/h, convert to m/s before substituting. This is a common error source in board exams.
Common Mistakes to Avoid
- Mistake 1 — Using g as positive in vertical equations. Gravity acts downward. In the standard sign convention (upward positive), always use \( -g \) in vertical equations. Many students write \( y = u_y t + \frac{1}{2}gt^2 \) instead of \( y = u_y t – \frac{1}{2}gt^2 \), leading to completely wrong answers.
- Mistake 2 — Forgetting that horizontal velocity is constant. There is no horizontal acceleration in ideal projectile motion. Students sometimes apply \( v = u + at \) horizontally with a non-zero value of a, which is incorrect.
- Mistake 3 — Confusing time to reach maximum height with total time of flight. The time to reach the top is \( t_{top} = \frac{u\sin\theta}{g} \), which is exactly half the total time of flight T. These two quantities are often swapped in exam answers.
- Mistake 4 — Applying the range formula for non-level ground. The formula \( R = \frac{u^2\sin 2\theta}{g} \) is valid only when the launch and landing points are at the same height. If the projectile lands on a cliff or in a valley, you must use the full kinematic equations.
- Mistake 5 — Incorrect angle for maximum range. Maximum range occurs at θ = 45°, not 60° or 30°. Confusing maximum range with maximum height (which occurs at θ = 90°) is a very common conceptual error.
JEE/NEET Application of Projectile Motion Formula
In our experience, JEE aspirants encounter projectile motion in 1–2 questions per paper, often combined with concepts from relative motion, circular motion, or energy conservation. NEET typically tests the formula in straightforward calculation problems worth 4 marks each.
Pattern 1 — Complementary Angles (JEE Main favourite): Questions ask students to prove or use the result that two angles of projection give the same range. The key relation is \( R_1 = R_2 \) when \( \theta_2 = 90^\circ – \theta_1 \). A related result often tested is \( H_1 + H_2 = \frac{R_{max}}{2} \) when \( \theta_1 + \theta_2 = 90^\circ \).
Pattern 2 — Projectile from a Height (JEE Advanced): A projectile is launched horizontally (\( \theta = 0 \)) from a height h. Here \( u_y = 0 \), time of fall is \( t = \sqrt{\frac{2h}{g}} \), and horizontal range is \( R = u\sqrt{\frac{2h}{g}} \). This variant is not directly covered by the standard formula and requires going back to first principles.
Pattern 3 — Velocity Direction at a Specific Point (NEET favourite): Questions ask for the angle the velocity vector makes with the horizontal after a given time. The approach is to find \( v_x = u\cos\theta \) and \( v_y = u\sin\theta – gt \), then compute \( \phi = \tan^{-1}(v_y / v_x) \). A special case tested in NEET is finding the time when the velocity is perpendicular to the initial velocity, which gives \( t = \frac{u}{g\sin\theta} \).
We recommend that JEE/NEET aspirants also study the concept of the “range on an inclined plane” and projectile motion in the presence of wind, as these appear in JEE Advanced papers and require extending the basic Projectile Motion Formula.
FAQs on Projectile Motion Formula
Explore more related topics on ncertbooks.net: visit our Physics Formulas hub for a complete collection. You may also find these articles useful: Critical Velocity Formula for circular motion concepts, Induced Voltage Formula for electromagnetic induction, and Superposition Formula for wave mechanics. For the official NCERT syllabus reference, visit ncert.nic.in.