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Motion in a Plane is the first chapter where a NEET(UG) question stops being about one straight line. A ball thrown at an angle, a swimmer crossing a river, a stone whirled on a string — all of them need you to break motion into two directions and treat each one separately. Students lose marks here not because the physics is hard, but because they forget that the horizontal and the vertical never talk to each other. Ten questions below. Give them ten honest minutes.
What this chapter carries in NEET(UG)
- Usually 2 to 3 questions every year, which is 8 to 12 marks.
- Most of them come from projectile motion and from adding two vectors.
- The rest come from relative velocity and from uniform circular motion.
What the syllabus says
In the syllabus (Unit 2, Kinematics): Motion in a plane, projectile motion, uniform circular motion.
Taken from the NMC syllabus for NEET (UG) 2026, the latest one published. See the full syllabus, unit by unit.
Formulas you must know by heart
| Time of flight | T = 2u sinθ / g |
| Maximum height | H = u2 sin2θ / 2g |
| Horizontal range | R = u2 sin2θ / g |
| Maximum range | Rmax = u2/g, at θ = 45° |
| Resultant of two vectors | R = √(A2 + B2 + 2AB cosθ) |
| Direction of the resultant | tanα = B sinθ / (A + B cosθ) |
| Centripetal acceleration | a = v2/r = ω2r |
| Relative velocity | vAB = vA − vB |
| Range equal to height | tanθ = 4 |
Look at the range formula once more. Because sin2θ is the same for θ and for 90° − θ, a ball thrown at 30° and a ball thrown at 60° with the same speed land at exactly the same place. NEET(UG) has asked that one more than once.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. A body is projected with speed u at an angle θ to the horizontal. Its maximum height is
- (a) u2sin2θ/g
- (b) u2sin2θ/2g
- (c) u2sin2θ/g
- (d) u2/2g
Q2. Two balls are thrown with the same speed, one at 30° and the other at 60°. The ratio of their horizontal ranges is
- (a) 1 : 1
- (b) 1 : 2
- (c) 2 : 1
- (d) 1 : 3
Q3. At the highest point of a projectile path, which quantity becomes zero?
- (a) The velocity
- (b) The vertical component of velocity
- (c) The acceleration
- (d) The horizontal component of velocity
Q4. Two vectors have magnitudes 3 units and 4 units. Which of these can never be the magnitude of their resultant?
- (a) 1 unit
- (b) 5 units
- (c) 7 units
- (d) 8 units
Q5. A particle moves in a circle of radius R with constant speed v. After half a revolution, the magnitude of the change in its velocity is
- (a) zero
- (b) v
- (c) 2v
- (d) v√2
Q6. A ball is projected at 45° with a speed of 20 m/s. Taking g = 10 m/s2, its time of flight is about
- (a) 1.41 s
- (b) 2.83 s
- (c) 2.00 s
- (d) 4.00 s
Q7. A river 40 m wide flows at 3 m/s. A swimmer whose speed in still water is 5 m/s heads straight across. The time taken to reach the other bank is
- (a) 8 s
- (b) 10 s
- (c) 13.3 s
- (d) 5 s
Q8. If |A + B| = |A − B| for two non-zero vectors, the angle between them is
- (a) 0°
- (b) 45°
- (c) 90°
- (d) 180°
Q9. A particle moves in a circle of radius 2 m with a constant speed of 4 m/s. Its centripetal acceleration is
- (a) 2 m/s2
- (b) 4 m/s2
- (c) 8 m/s2
- (d) 16 m/s2
Q10. For a projectile, the horizontal range is equal to the maximum height. The angle of projection satisfies
- (a) tanθ = 1
- (b) tanθ = 2
- (c) tanθ = 4
- (d) tanθ = 1/4
Answer key, with a hint for each
| Q1 | (b) | Split u into u cosθ and u sinθ. Only the vertical part climbs, and it dies out at height u2sin2θ/2g. |
| Q2 | (a) | sin2θ is the same for 30° and 60°, because sin60° = sin120°. Same speed, same range. |
| Q3 | (b) | The ball still moves forward at the top. It is only the upward part of the velocity that has been eaten by gravity. |
| Q4 | (d) | A resultant can only lie between |A − B| and A + B, that is between 1 and 7. Eight is outside that window. |
| Q5 | (c) | Speed is the same but the direction has reversed, so the change is v − (−v) in size, which is 2v. |
| Q6 | (b) | T = 2u sinθ/g = 2 × 20 × 0.707 / 10. |
| Q7 | (a) | The crossing time depends only on the part of the speed that points across the river. The current carries the swimmer downstream but does not delay the crossing. |
| Q8 | (c) | Square both sides. The cross term 2AB cosθ must vanish, so cosθ = 0. |
| Q9 | (c) | a = v2/r = 16/2. |
| Q10 | (c) | Put R = H. u2sin2θ/g = u2sin2θ/2g gives 4cosθ = sinθ. |
Now mark yourself honestly
NEET(UG) marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 45 Physics questions are out of 180.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in May.
The full chapter, on a clock
Ten questions tell you a little. Forty-five tell you the truth.
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