Projectile Motion
Introduction
When a cricketer lofts a six, when Neeraj Chopra's javelin sails to 87.58 m at the Tokyo Olympics, or when DRDO test-fires an Agni missile, the same physics is at work: projectile motion. The good news for aspirants is that this topic is small, formula-driven and predictable. If you are preparing for SSC, UPSC, NDA, CDS, RRB or a State PSC while juggling a job or college, this guide gives you the concepts, the shortcuts, the traps examiners love, and a practice set, all in one sitting.
Look for the Exam Notes boxes. The points that actually show up in papers.
What Is Projectile Motion?
Projectile motion is the motion of an object thrown into the air that moves
under the influence of gravity alone. The object is called a projectile, and the
curved path it follows is its trajectory.
A few terms you must know:
- Point of projection: where the object is launched from.
- Angle of projection (θ): the angle between the launch direction and the horizontal.
- Time of flight (T): total time the object stays in the air.
- Range (R): horizontal distance covered.
- Maximum height (H): the highest vertical point reached.
Exam questions assume three things unless told otherwise: no air resistance, constant gravity (g = 9.8 m/s², often rounded to 10), and a short enough range that the Earth can be treated as flat.
The one idea that unlocks everything: horizontal and vertical motions are independent. Horizontally, nothing speeds the object up or slows it down, so velocity stays constant. Vertically, gravity pulls it down at g, exactly like a ball thrown straight up. Once you split every problem into these two parts, the formulas stop feeling like things to memorize.
Exam Note: The path of a projectile is a parabola. This is
asked directly in SSC and railway exams.
Periodic Classification of Elements: Notes & MCQ.
Types of Projectile Motion
1. Oblique (angular) projectile. Launched at an angle θ to the horizontal. A football kick, a cricket shot or a javelin throw all fit here. This is the type most questions use.
2. Horizontal projectile. Launched horizontally from a height, so the initial vertical velocity is zero. Think of a ball rolling off a table or a food packet dropped from a plane flying level.
3. Projectile on an inclined plane. The ground is a slope instead of flat. It is mostly relevant for JE and other technical exams, so general-stream aspirants can skip it.
Projectile
Motion Formulas
Let u be the launch speed and θ
the angle of projection.
|
Quantity |
Formula |
What it means |
|
Horizontal component |
u cosθ |
Stays constant throughout |
|
Vertical component |
u sinθ |
Changes at rate g |
|
Time of flight |
T = 2u sinθ / g |
Time up plus time down |
|
Maximum height |
H = u² sin²θ / 2g |
Depends only on the vertical part |
|
Range |
R = u² sin2θ / g |
Horizontal distance on level ground |
|
Maximum range |
R = u² / g |
Happens at θ = 45° |
|
Trajectory equation |
y = x tanθ − gx² / (2u² cos²θ) |
Proves the path is a parabola |
|
Horizontal projection from height h |
T = √(2h/g), R = u√(2h/g) |
Fall time depends only on height |
How to remember them:
- Time has sinθ (vertical stuff). Range has sin2θ. Height has sin²θ.
- For T, think "go up for u sinθ/g, come down for the same time, so double it."
- Height and time depend only on the vertical component. Only range brings in the horizontal.
Exam Note: In SSC
CGL and CHSL, most questions are a direct plug-in of one of these formulas.
Memorize T, H and R first.
Animal Kingdom Complete Classification: Notes & MCQ.
Shortcuts and Key Relations for Quick Solving
These save 30 to 40 seconds per
question, which matters when you have under a minute for each one.
1. Complementary angles give the same range. Angles θ and (90° − θ) have equal range because sin2θ and sin(180° − 2θ) are equal. So 30° and 60° give the same range, and so do 15° and 75°.
2. Range is maximum at 45°, height is maximum at 90°. At 90° the object goes straight up, so it has height but no range.
3. R = 4H cotθ. This comes from dividing R by H. At 45°, cot45° = 1, so R = 4H. If a question says the range is 80 m at 45°, the maximum height is 20 m. No calculation needed.
4. Scaling with speed. Double the launch speed at a fixed angle, and T doubles, while H and R become four times. Triple the speed, and H and R become nine times.
5. Symmetry. When launch and landing are at the same level, time up equals time down, and speed at any height is the same on the way up and on the way down.
Worked shortcut: A ball is thrown at 45° and lands 100 m away. At 15°, the same speed gives a range of 100 × sin30° = 50 m. You skip finding u entirely, because R = (u²/g) × sin2θ and u²/g is just the maximum range.
Exam Note: "Ratio of range to maximum
height" and "same range at complementary angles" are favourite
shortcut-style questions. Learn R = 4H cotθ cold.
Supreme Court of India — the complete UPSC/SSC notes.
Common Mistakes and Traps in Exams
Examiners know where students slip. Most conceptual MCQs test one of these.
|
Trap |
Truth |
|
"Acceleration is zero at the top." |
Acceleration is g downward at every point, including the top. |
|
"Velocity is zero at the top." |
Only the vertical component is zero. The horizontal component u cosθ remains. |
|
"Horizontal velocity decreases as the ball rises." |
It stays constant (no air resistance). |
|
"A heavier object travels farther." |
Mass does not affect the trajectory in the ideal case. |
|
"Range and displacement are the same." |
Range is horizontal distance. Displacement is the straight-line distance from start to end, and it differs unless the object lands at the same level. |
|
"Speed and velocity at the top are zero." |
Speed at the top is u cosθ, which is the minimum speed along the path. |
Exam Note: The
question "What is the acceleration of a projectile at its highest
point?" has appeared in many forms across exams. The answer is always g,
directed downward.
483 concepts students know—but still get wrong.
Quick Revision Box
- Path is a parabola. Horizontal velocity is constant, vertical velocity changes at g.
- T = 2u sinθ / g, H = u² sin²θ / 2g, R = u² sin2θ / g.
- Maximum range = u²/g at 45°. Maximum height at 90°.
- θ and (90° − θ) give the same range.
- R = 4H cotθ, so R = 4H at 45°.
- Acceleration is g downward everywhere, including the top.
- At the top, velocity = u cosθ (not zero), the minimum speed.
- Double the speed: T doubles, H and R become 4 times.
- Horizontal projection: T = √(2h/g), R = u × T.
- Mass does not affect the trajectory. Air resistance is ignored in exams.
MCQs
Q1. If a projectile is thrown with velocity v at an angle θ with the x-axis, the time taken to reach maximum height is given by:
(a) t = v sinθ / g
(b) t = v² sinθ / g
(c) t = v² sin²θ / g
(d) t = v sin²θ / g
Answer: (a)
Explanation: At the highest point, the vertical velocity is zero. The vertical component starts at v sinθ and falls at rate g, so 0 = v sinθ − gt, which gives t = v sinθ / g. This is half the total time of flight (2v sinθ / g).
Reference: SSC Scientific Assistant (Physics), 22 Nov 2017, Shift 2
Q2. The range of a projectile is maximum when the angle of projection is:
(a) 60°
(b) 45°
(c) 90°
(d) 30°
Answer: (b)
Explanation: Range R = u² sin2θ / g. It is largest when sin2θ = 1, which means 2θ = 90°, so θ = 45°. The maximum range is u²/g.
Reference: RRB JE CBT 2 (Mechanical), 19 Sep 2019, Shift 2
Q3. The maximum height of a projectile is given by:
(a) u² sin²α / 2g
(b) u² cos²α / 2g
(c) u² sin2α / g
(d) u² cos2α / 2g
Answer: (a)
Explanation: Height depends only on the vertical component, u sinα. Using v² = u² − 2gh with v = 0 at the top gives H = (u sinα)² / 2g. Option (c) is the formula for range, a common mix-up.
Reference: RRB JE (Civil) CBT 2, 28 Aug 2019, Shift 2
Q4. In projectile motion, where is the kinetic energy minimum?
(a) At the point of projection
(b) At the highest point
(c) At the point of landing
(d) At the midpoint of the range
Answer: (b)
Explanation: Kinetic energy is ½mv². Speed is lowest at the top, where only the horizontal component u cosθ remains. Speed (and KE) is therefore minimum but not zero at the highest point. This is a classic trap.
Reference: RRB JE (IT) CBT II, 30 Aug 2019
Q5. The path followed by a projectile is called its:
(a) Range
(b) Trajectory
(c) Displacement
(d) Amplitude
Answer: (b)
Explanation: The curved path of a projectile is its trajectory, and for ideal projectile motion it is a parabola. Range is only the horizontal distance covered.
Reference: RRB JE CBT I, 2 June 2019, Shift 2
Q6. What should be the angle of throw of a projectile to achieve maximum vertical range?
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Answer: (c)
Explanation: Vertical range means maximum height, H = u² sin²θ / 2g. It is largest when sinθ = 1, so θ = 90° (a straight vertical throw). Maximum horizontal range is at 45°, so read the question carefully.
Reference: IB ACIO Grade-II, Paper I, 15 Sep 2013
Q7. The horizontal range of a projectile is 4√3 times its maximum height. The angle of projection is:
(a) 15°
(b) 45°
(c) 30°
(d) 60°
Answer: (c)
Explanation: Use R = 4H cotθ. Since R = 4√3 H, we get cotθ = √3, so θ = 30°. Without the shortcut, divide R by H and simplify, which takes much longer.
Reference: HTET TGT Science 2020
Q8. A projectile is fired at an angle θ to the vertical. Its horizontal range will be maximum when θ is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b)
Explanation: The angle with the horizontal is (90° − θ). Range is maximum when that angle is 45°, so θ = 45° as well. This works only because 45° is its own complement, so the answer is the same whether measured from the horizontal or the vertical. For other angles, you’d have to convert.
Reference: ISRO VSSC Technical Assistant (Mechanical), 8 Feb 2015
Q9. The time of flight of a projectile on a horizontal plane, with initial velocity u, angle θ and gravitational acceleration g, is:
(a) u sinθ / g
(b) 2u sinθ / g
(c) u² sin2θ / g
(d) u² sin²θ / 2g
Answer: (b)
Explanation: Time up equals u sinθ / g, and time down is the same for a level landing, so total T = 2u sinθ / g. Option (a) is only the time to the top, (c) is the range and (d) is the height.
Reference: JKSSB JE (Mechanical), 2015
Q10. Which of the following remains constant for a projectile fired from the earth?
(a) Speed
(b) Vertical component of velocity
(c) Horizontal component of velocity
(d) Kinetic energy
Answer: (c)
Explanation: With no air resistance, nothing acts horizontally, so u cosθ never changes. Gravity acts vertically, so the vertical component, the speed and the kinetic energy all change.
Reference: UP TGT Science 2011 (held in 2016)
Colonial Economy – one of the most important topics for History section.
Conclusion:
Projectile
motion looks intimidating only until you split it into a constant horizontal
motion and a gravity-driven vertical motion. Learn the three core formulas, the
45° and complementary-angle shortcuts, and the "top of the path"
traps, and you can handle nearly everything these exams throw at you.
If you study after work, do this in three short sessions: formulas on day one, shortcuts and traps on day two, and the practice questions on day three. Then take a physics mock test to check how well it stuck.
Frequently Asked Questions
At what
angle is the range of a projectile maximum?
The range is maximum at 45° when launch and landing are at the same level. The maximum range equals u²/g. At complementary angles, such as 30° and 60°, the range is the same.
What is the
acceleration of a projectile at its highest point?
It is g, about 9.8 m/s², directed downward. Gravity acts throughout the flight, so acceleration never becomes zero. Only the vertical velocity is zero at the top.
What is the
shape of a projectile's path?
A parabola. The trajectory equation y = x tanθ − gx² / (2u² cos²θ) is quadratic in x, which is the equation of a parabola.
What is the
difference between projectile motion and free fall?
In free fall, an object starts from rest or moves vertically under gravity. A projectile also has a horizontal velocity. Free fall is projectile motion with zero horizontal speed.

