Motion in a Plane
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Direct answer
Projectile motion superposes two independent motions: a constant horizontal velocity u cosθ and a vertical free fall, and every standard result — time of flight, maximum height, range — falls out of that split. The chapter also supplies the vector algebra toolkit (dot and cross products) and uniform circular motion, where the speed is constant yet the acceleration v^2/r points steadily to the centre. River-crossing problems complete the NEET-UG set by combining vectors with relative velocity.
What you must remember
- Projectile launched at angle θ with speed u: time of flight T = 2u sinθ/g, maximum height H = u^2 sin^2θ/2g, horizontal range R = u^2 sin2θ/g.
- Range is maximum at θ = 45°; complementary angles θ and 90° - θ give identical ranges, so 30° and 60° share one value of R.
- At the top of the trajectory only the vertical component dies: the velocity there is u cosθ, horizontal, not zero.
- Body projected horizontally from height h: time to ground t = sqrt(2h/g), independent of launch speed; range = u sqrt(2h/g).
- Dot product A·B = AB cosθ is a scalar and commutative; cross product magnitude |A × B| = AB sinθ is a vector along the right-hand-rule normal and reverses sign on swapping.
- Uniform circular motion: centripetal acceleration a = v^2/r directed radially inward; the tangential speed never changes.
- River crossing: shortest time comes from heading straight across (t = d/v_boat, accepting drift); zero drift requires the boat aimed upstream at an angle whose sine is v_river/v_boat.
Worked example: the 30° launch
A projectile leaves the ground at 20 m/s at 30°. Resolve once, carefully: u_x = 20 cos30° = 17.3 m/s, u_y = 20 sin30° = 10 m/s. Time of flight T = 2 × 10/9.8 ≈ 2.04 s. Maximum height H = 100/(2 × 9.8) ≈ 5.1 m. Range R = u_x × T = 17.3 × 2.04 ≈ 35.3 m, which the formula u^2 sin60°/9.8 confirms. Now the comparison step that examiners love: a 60° launch at the same speed has the same range (sin120° = sin60°) but twice the height and a longer flight. If the question asks for the speed at the top, the answer is simply 17.3 m/s — the surviving horizontal component; if it asks for the velocity direction at the top, it is horizontal, because gravity has momentarily exhausted the vertical component.
Extend the same discipline to a horizontal launch from a 20 m cliff at 15 m/s: the fall time is fixed by the cliff, t = sqrt(2 × 20/9.8) ≈ 2.02 s, and the range is 15 × 2.02 ≈ 30.4 m. Speed matters only along one axis at a time — that is the entire method.
Where students slip
The recurring error is smuggling a horizontal acceleration into the problem: with air resistance neglected, gravity acts only vertically, so horizontal velocity never changes and the trajectory is parabolic for that reason alone. The second error is declaring the projectile "momentarily at rest" at the apex; the vertical component is zero there, but u cosθ marches on, which is precisely why the path flattens rather than corners at the top. Third, students apply v = u + at as a scalar to the whole speed instead of working per component — the equations of motion are three separate statements, one per axis. In circular motion, the claim "acceleration is zero because speed is constant" is the standard distractor; the direction of the velocity vector is changing every instant, and that changing direction is itself acceleration, v^2/r. Finally, in cross-product questions the answer's direction is asked as often as its magnitude — the right-hand rule decides it, and reversing the order of the vectors reverses the answer.
Frequently asked questions
Why is the horizontal velocity of a projectile constant?
No horizontal force acts on it (air resistance neglected), so by Newton's first law the horizontal component u cosθ is unchanged throughout the flight.
Which two launch angles give the same horizontal range?
θ and 90° - θ, because sin2θ = sin(180° - 2θ). Hence 20° and 70°, or 30° and 60°, are range-twins; 45° stands alone with the maximum.
What supplies the acceleration in uniform circular motion?
A real force — tension, friction, gravity or a normal reaction — directed along the radius toward the centre, of magnitude v^2/r. There is no outward "centrifugal" force in the ground frame.
When is the magnitude of the cross product of two vectors maximum?
When the vectors are perpendicular, giving |A × B| = AB; it is zero for parallel (or antiparallel) vectors. The dot product behaves oppositely.
How should a boat head to cross a river in minimum time?
Straight across the bank. The crossing time is the river's width divided by the boat's own speed; any upstream angle shortens the across-component and lengthens the time.