# Projectile Range Numericals

> Range R = u² sin2θ/g with height, time of flight and complementary angle results for NEET Physics kinematics.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/projectile-range-numericals-neet
- Exam / course: NEET-UG · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Projectile Range Numericals", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/projectile-range-numericals-neet

## Direct answer

Forty-five degrees is not superstition: range R = u² sin 2θ/g peaks when sin 2θ = 1, and every complementary pair (30° with 60°, 20° with 70°) delivers the same range because sin 2θ repeats under θ → 90° − θ. The complete kit for ground-to-ground flight: time of flight T = 2u sinθ/g, maximum height H = u² sin²θ/2g, range R = u² sin 2θ/g, and the parabolic trajectory y = x tanθ − gx²/(2u² cos²θ). Underneath runs the split that solves everything — uniform velocity u cosθ horizontally, uniform acceleration g vertically, two independent 1-D problems sharing only the clock.

## What you must remember

- **The three formulas:** R = u² sin2θ/g, H = u² sin²θ/2g, T = 2u sinθ/g — valid for launch and landing at the same level.
- **Complementary angle fact:** θ and 90° − θ give equal range; the higher angle stays airborne longer and rises more (60° reaches three times the height of 30°, from sin²60°/sin²30° = 3).
- **Maximum range:** R_max = u²/g at 45°; no other single angle does better.
- **Symmetry facts:** speed at landing equals speed at launch (same level); at half the time the projectile is at H; at half the range it is at its peak with purely horizontal velocity u cosθ.
- **Trajectory equation:** y = x tanθ − gx²/(2u²cos²θ) — a parabola, obtained by eliminating t between the two 1-D equations.
- **Horizontal launch from height h:** time t = √(2h/g) independent of launch speed; range = u√(2h/g).
- **Velocity components:** v_x = u cosθ always; v_y = u sinθ − gt; the negative sign in v_y is where most sign errors enter.

## Running the full numerical

Launch at u = 20 m/s, θ = 30°, g = 10 m/s². Time of flight: T = 2 × 20 × 0.5/10 = 2 s. Height: H = 400 × 0.25/20 = 5 m. Range: R = 400 × sin 60°/10 = 400 × 0.866/10 ≈ 34.6 m. Repeat at 60°: same range, but T = 3.46 s and H = 15 m — the complementary-angle contract honoured in numbers. Now the cliff variant: throw horizontally at 20 m/s from 45 m. Fall time from height alone: t = √(2 × 45/10) = 3 s; range = 60 m; vertical velocity on impact = gt = 30 m/s; impact speed = √(20² + 30²) = 36 m/s at tan⁻¹(3/2) ≈ 56° below horizontal. One problem, every formula, no memorisation beyond the three lines.

## Where NEET sets the trap

The calculator-level trap is sin 2θ entered as 2 sinθ — at θ = 30° that swaps 0.866 for 1.0, and both resulting ranges appear in the options. The conceptual trap is the symmetric-pair claim tested backwards: given that 20° gives range R, which other angle matches it? Seventy degrees, not forty. Motion-division traps ask for velocity at the highest point (purely horizontal, u cosθ, not zero) or the average velocity over the whole flight — for same-level launches that equals displacement over time, a horizontal u cosθ, a result almost no student derives but the paper occasionally asks. Cliff problems punish those who compute fall time from the projectile formulas instead of √(2h/g). Air resistance is ignored throughout, matching NCERT's idealisation, so never "correct" for it.

## Frequently asked questions

### At what angle of projection is the horizontal range maximum?

45°, where sin 2θ = 1 and R_max = u²/g; angles equally above and below 45° share the same range.

### Why do 30° and 60° launches land at the same spot?

Because sin 2θ = sin(180° − 2θ): sin 60° = sin 120°; their maximum heights differ, with 60° rising three times higher than 30°.

### For u = 20 m/s at 30° with g = 10 m/s², what are T, H and R?

T = 2 s, H = 5 m and R ≈ 34.6 m, by direct substitution into 2u sinθ/g, u²sin²θ/2g and u²sin2θ/g.

### What is the velocity of a projectile at its highest point?

Purely horizontal with magnitude u cosθ; the vertical component vanishes there while the acceleration remains g downward.

### A stone thrown horizontally at 20 m/s from a 45 m cliff lands when?

After t = √(2h/g) = 3 s, independent of the 20 m/s; it strikes about 60 m from the cliff foot at 36 m/s.