Projectile Motion — Trajectory, Time of Flight, Height, Range (NCERT 3.9)
1. What Is Projectile Motion?
A projectile is any body thrown into the air and thereafter governed only by gravity — a cricket ball at 30°, a stone kicked off a cliff horizontally, a long jumper after take-off. The physics is entirely the constant-acceleration machinery of the previous topic with one fixed choice of axes: x horizontal, y vertical, a = −g ĵ. Two modelling assumptions make every formula in this page valid: gravity is uniform (g = 9.8 m s⁻², near the surface), and air resistance is ignored. Drop either assumption and the parabola bends or shortens — but NTA's questions live inside these assumptions.
The strategic key is independence of the two motions: the horizontal world has no force, so vx never changes; the vertical world is plain free fall. The ball's horizontal velocity at landing equals its horizontal velocity at launch — a fact that feels wrong to beginners and is therefore NTA's favourite assertion.
2. Complete Theory: Every Formula, Derived
Components of launch. Speed u at angle θ above the horizontal: vx = u cos θ (constant forever) and vy = u sin θ − gt (falls linearly, crosses zero at the top, reverses). The position equations are x = (u cos θ)t and y = (u sin θ)t − ½gt².
Time of flight T (launch and landing at the same height): set y = 0 at t > 0 → (u sin θ)t = ½gt² → T = 2u sin θ/g. Only the vertical story decides T — the ball cannot know how fast it moves sideways.
Maximum height H: at the top vy = 0, so by v² = u² + 2as applied vertically, 0 = u²sin²θ − 2gH → H = u² sin²θ/(2g). Reached at t = u sin θ/g — exactly half of T, since the vertical motion is symmetric.
Horizontal range R: x at t = T: R = (u cos θ)(2u sin θ/g) = u² sin 2θ/g, using 2 sin θ cos θ = sin 2θ. Since sin 2θ peaks at 2θ = 90°, R is maximum at θ = 45°: Rmax = u²/g (same-height condition!). And because sin 2(90° − θ) = sin 2θ, complementary angles (θ, 90° − θ) share the same range with different H and T.
Trajectory equation: eliminate t from x = (u cos θ)t → y = x tan θ − gx²/(2u²cos²θ) — a parabola in x, as promised. Setting y = 0 recovers R (self-check); setting dy/dx = 0 recovers x = R/2 at the apex (another self-check).
Horizontal projection from a height h (cliff, table, arrow off a wall): vertical motion is free fall from rest → t = √(2h/g) (independent of u!), horizontal reach x = u√(2h/g), and the impact speed by v² = 2gh vertically plus u² horizontally: v = √(u² + 2gh) at tan⁻¹(√(2gh)/u) below the horizontal.
Velocity at any instant: v = √(vx² + vy²) = √(u² − 2gy) — a neat form showing the speed depends only on the height, so the landing speed equals the launch speed (energy conservation previewing Chapter 6), and speeds at equal heights are equal.
3. Visualising the Parabola
Figure 3.8 — Anatomy of a same-level projectile
Orange arrows are velocities: tilted at launch, purely horizontal at the apex, tilted (mirror image) at landing.
Exam read-out: the launch and landing velocity arrows are mirror images (same speed u, angles ±θ) — the vertical motion is time-symmetric about the apex. Symmetry instantly gives: tup = tdown = T/2, and speeds at equal heights are equal.
4. Solved Examples
Example 1 — The full formula set (u = 20 m/s, θ = 30°, g = 10 m/s²)
A ball is kicked at 20 m/s, 30° above the horizontal on level ground. Find T, H, R, and the velocity at t = 1 s.
Solution (step by step): ux = 20 cos 30° = 10√3 ≈ 17.32 m/s; uy = 20 sin 30° = 10 m/s. (i) T = 2uy/g = 2(10)/10 = 2 s. (ii) H = uy²/2g = 100/20 = 5 m. (iii) R = uxT = 17.32 × 2 ≈ 34.6 m (= u² sin 60°/g = 400 × 0.866/10 ✓). (iv) At t = 1 s (the apex): v = ux = 10√3 ≈ 17.3 m/s horizontal — vy = 10 − 10(1) = 0 exactly. Sanity chain: H also equals uyT/4 = 10×2/4 = 5 ✓.
Example 2 — Horizontal projection from a cliff (u = 15 m/s, h = 20 m, g = 10)
A stone is kicked horizontally at 15 m/s from a 20 m cliff. Find the flight time, the landing distance from the base, and the impact velocity.
Solution: (i) Vertical free fall from rest: t = √(2h/g) = √(40/10)... = √4 = 2 s — independent of the 15 m/s. (ii) Horizontal reach: x = u t = 15 × 2 = 30 m. (iii) Impact: vx = 15, vy = gt = 20 m/s downward → v = √(225 + 400) = √625 = 25 m/s, at tan⁻¹(20/15) = 53.1° below the horizontal. The 15-20-25 triangle (3-4-5 family) is no accident — NTA picks launch data to keep impact speeds exact.
5. Practice Questions
Q1. For u = 20 m/s at 45° (g = 10 m/s²), find T, H and R.
ANSWER: u_y = 20 × 0.707 = 14.14 m/s. T = 2(14.14)/10 ≈ 2.83 s. H = (14.14)²/20 = 200/20 = 10 m. R = u²sin 90°/g = 400/10 = 40 m — the maximum range for this u, at the 45° champion angle.
Q2. Show that projectiles launched at 30° and 60° with the same speed land at the same point, and state which one stays airborne longer.
ANSWER: sin 2(60°) = sin 120° = sin 60° = sin 2(30°) → identical R. Time of flight T = 2u sin θ/g: sin 60° > sin 30°, so the 60° shot stays in the air √3 times longer (and rises 3× higher, H ∝ sin²θ).
Q3 (MCQ). A ball thrown from level ground takes 4 s to return. If g = 10 m/s², its launch speed and max height are: (a) 20 m/s, 20 m (b) 40 m/s, 40 m (c) 20 m/s, 10 m (d) 10 m/s, 5 m
ANSWER: (a). T = 2u sin θ/g = 4 → u sin θ = 20 m/s; H = (u sin θ)²/2g = 400/20 = 20 m. Note only u sin θ is fixed — the horizontal component (and hence u) needs range data too.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| vx = u cos θ (constant); vy = u sin θ − gt | No air resistance; up positive |
| T = 2u sin θ/g | Same launch/landing height; vertical story only |
| H = u² sin²θ/(2g) | Apex at t = T/2; height champion is θ = 90° |
| R = u² sin 2θ/g; Rmax = u²/g at 45° | Same height; complementary angles tie |
| y = x tan θ − gx²/(2u²cos²θ) | Parabola; y = 0 recovers R |
| Horizontal launch: t = √(2h/g); x = u√(2h/g) | Flight time independent of u |
| v(t) = √(u² − 2gy) | Speed depends on height only; landing speed = u |
| [R] = L, [T] = T, [H] = L | Dimensional self-check à la Ch 1 |
7. Frequently Asked Questions
What is the velocity of a projectile at the highest point of its path?
At the top, the vertical component v_y = u sin θ − gt has fallen to zero, but the horizontal component v_x = u cos θ is untouched because no horizontal force acts. The velocity at the highest point is therefore u cos θ, entirely horizontal, and the acceleration there is g, vertically downward — not zero. Only for a purely vertical throw (θ = 90°) does the speed become zero at the top.
Why do two complementary launch angles give the same horizontal range?
The range R = u² sin 2θ/g depends on sin 2θ, and sin 2(90° − θ) = sin(180° − 2θ) = sin 2θ — identical. So θ and 90° − θ always land at the same spot for the same u. The low angle reaches the target faster (shorter time of flight, smaller maximum height), while the high angle hangs in the air longer; a common MCQ asks exactly this distinction.
At what launch angle is the horizontal range maximum?
R = u² sin 2θ/g is maximum when sin 2θ = 1, i.e., 2θ = 90° or θ = 45°, giving R_max = u²/g. This maximum requires launch and landing at the same height; from a raised platform or cliff the optimal angle drops below 45°, which is why the same-height condition is printed alongside every 45° statement.
How is the trajectory equation of a projectile derived?
Write the two motion equations: x = (u cos θ)t and y = (u sin θ)t − ½gt². Eliminate t by substituting t = x/(u cos θ) into the y-equation: y = x tan θ − gx²/(2u² cos²θ). Since tan θ and 1/cos²θ are constants for fixed u and θ, y is quadratic in x — a parabola. Setting y = 0 recovers the range x = u² sin 2θ/g, a built-in self-check of the derivation.
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