Projectile motion
Fire a ball at any speed and angle and watch it fly the exact parabola the equations predict. So launch it and watch. Change the speed or the angle and the time of flight, max height and range all update instantly — the same v₀, θ, g in the readout as on the canvas.
Using g = 9.8 m/s² (fixed) — no air resistance.
Simplified model for intuition — NCERT's formulas above assume no air resistance.
The physics you're seeing
Horizontal motion
No horizontal force acts on the ball once it leaves your hand (we ignore air resistance), so the horizontal velocity vx = v₀cosθ stays exactly constant for the whole flight.
Vertical motion
Gravity pulls down the entire time, so the vertical velocity changes uniformly: vy = v₀sinθ − gt. It's plain 1D motion under gravity, riding along with the constant horizontal drift.
Time of flight
T = 2v₀sinθ/g. Complementary angles land at the same spot: sin(2×30°) = sin(60°) = sin(120°) = sin(2×60°), so 30° and 60° launches (same speed) share a range — just very different flight times.
Maximum range
R = v₀²sin(2θ)/g is largest when sin(2θ) = 1 — that's θ = 45°. For a fixed launch speed, any angle steeper or flatter than 45° falls short of it.
Part of the Motion in a Plane chapter — read the notes, grab the formula sheet and take the quiz. One of Priodemy for School, free with every EduSuite school.
Two independent motions, happening at once
The idea that makes it easy
A projectile looks complicated because the path is curved, but the curve is the result of two very simple motions running side by side and not affecting each other at all. Horizontally there is no force (ignoring air resistance), so the horizontal velocity u cos θ never changes. Vertically gravity acts constantly downward, so the vertical velocity changes at 9.8 m/s every second, exactly as in free fall.
Once you accept the independence, every projectile question becomes two ordinary one-dimensional problems joined by a shared clock. Time is the only quantity the two directions have in common, which is why almost every solution starts by finding the time of flight.
A famous demonstration makes the point: a ball dropped from a height and a ball fired horizontally from the same height hit the ground simultaneously. The horizontal motion contributes nothing to the fall, however fast the second ball is travelling.
Where the standard results come from
At the top of the flight the vertical velocity is momentarily zero, which gives a time to the peak of u sin θ / g. The path is symmetric, so the total time of flight is T = 2u sin θ / g. Maximum height follows from the vertical motion alone: H = u²sin²θ / 2g.
Horizontal range is then just constant horizontal speed multiplied by that time, which simplifies to R = u²sin 2θ / g. Because sin 2θ peaks when 2θ = 90°, the maximum range occurs at 45° — provided launch and landing are at the same height. The sin 2θ form also explains why complementary angles give equal range: 30° and 60° land in the same place, though by very different paths, one flat and fast, the other high and slow. Fire both here and compare.
Mistakes that cost marks
Applying the 45° result when the heights differ. Launching from a cliff or throwing at a raised target breaks the symmetry, and the optimal angle is no longer 45°. The standard formulas assume equal launch and landing heights.
Thinking acceleration is zero at the top. The vertical velocity is zero there; the acceleration is still g, downward, throughout the entire flight. If it were zero the projectile would sail off horizontally.
Mixing the components in one equation. Resolve first, then work in each direction separately. Substituting the full launch speed u into a vertical equation, instead of u sin θ, is the most common source of wrong answers here.