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Motion in a Plane is where JEE Main starts testing whether you can actually handle vectors, not just name them. Projectiles given as î and ĵ components, river crossing, circular motion — all of them are easy once you break the motion into two independent directions, and all of them are a mess if you try to keep them in one. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 1 to 2 questions every year, which is 4 to 8 marks.
- Projectile motion in component form is the most common one.
- Relative velocity and uniform circular motion share the rest.
What the syllabus says
In the syllabus (Unit 2, Kinematics): Motion in a plane, projectile motion, uniform circular motion.
Taken out in 2024: the vector block — scalars and vectors, addition and subtraction, zero vector, scalar and vector products, unit vector, resolution of a vector.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Time of flight | T = 2u sinθ / g |
| Maximum height | H = u2sin2θ / 2g |
| Horizontal range | R = u2sin2θ / g |
| Range and height | R/H = 4 cotθ |
| Resultant of two vectors | R = √(A2 + B2 + 2AB cosθ) |
| Angular velocity | ω = 2πn |
| Centripetal acceleration | a = v2/r = ω2r |
| Projectile from a height | t = √(2h/g), horizontal distance = ut |
| River crossing, straight across | sinθ = vriver/vboat |
| Two products compared | |A × B| / (A·B) = tanθ |
The line R/H = 4 cotθ is worth keeping. It links the two most asked quantities of this chapter in one step, and at 45° it tells you at once that the range is four times the maximum height.
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. For a projectile thrown at 45°, the ratio of its range to its maximum height is
- (a) 2
- (b) 4
- (c) 1
- (d) 8
Q2. A particle is projected with velocity (3î + 4ĵ) m/s. Taking g = 10 m/s2, its time of flight is
- (a) 0.4 s
- (b) 0.8 s
- (c) 0.6 s
- (d) 1.0 s
Q3. Two vectors each of magnitude 5 units have a resultant of magnitude 5 units. The angle between them is
- (a) 60°
- (b) 90°
- (c) 120°
- (d) 150°
Q4. A boat whose speed in still water is 5 km/h must cross a river flowing at 3 km/h and land directly opposite. It should head, measured from the line straight across, at
- (a) 30°
- (b) 37°
- (c) 53°
- (d) 45°
Q5. In uniform circular motion, the angle between the velocity and the acceleration is
- (a) 0°
- (b) 45°
- (c) 90°
- (d) 180°
Q6. A projectile thrown at an angle θ has the same range as one thrown at
- (a) θ/2
- (b) 90° − θ
- (c) 2θ
- (d) 180° − θ
Q7. During the flight of a projectile, the horizontal component of its velocity
- (a) increases steadily
- (b) decreases steadily
- (c) stays constant
- (d) becomes zero at the top
Q8. A stone tied to a string 1 m long is whirled at 4 revolutions per second. Taking π2 = 10, its centripetal acceleration is
- (a) 64 m/s2
- (b) 160 m/s2
- (c) 640 m/s2
- (d) 320 m/s2
Q9. A body is thrown horizontally at 10 m/s from a height of 80 m. Taking g = 10 m/s2, it lands at a horizontal distance of
- (a) 20 m
- (b) 40 m
- (c) 80 m
- (d) 10 m
Q10. A body is projected with speed u at 45° to the horizontal. The ratio of its maximum height to its horizontal range is
- (a) 1 : 1
- (b) 1 : 2
- (c) 1 : 4
- (d) 2 : 1
Answer key, with a hint for each
| Q1 | (b) | R/H = 4cotθ, and cot 45° = 1. |
| Q2 | (b) | The vertical component is 4 m/s, so T = 2 × 4 / 10. |
| Q3 | (c) | 25 = 25 + 25 + 50cosθ gives cosθ = −1/2. |
| Q4 | (b) | The upstream component must cancel the current, so sinθ = 3/5, which is 37°. |
| Q5 | (c) | The speed never changes, so the acceleration can only turn the velocity. It must therefore act at right angles to it. |
| Q6 | (b) | sin2θ takes the same value for θ and for 90° − θ. |
| Q7 | (c) | There is no horizontal force, so there is no horizontal acceleration. |
| Q8 | (c) | ω = 2π × 4 = 8π, so a = ω2r = 64π2 = 640. |
| Q9 | (b) | It falls for t = √(2 × 80/10) = 4 s, and travels 10 × 4 horizontally in that time. |
| Q10 | (c) | H = u²sin²θ/2g and R = u²sin2θ/g. At 45° these are u²/4g and u²/g, so H : R = 1 : 4. The ratio is 1 : 4 only at 45° — at any other angle it changes. |
Now mark yourself honestly
JEE Main 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, 25 Physics questions are out of 100.
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 January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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