Projectile Range Numericals
On this page
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.