Projectile at an Angle: Master Trajectory Physics the Easy Way

JEE/NEET Physics · Motion in a Plane series · Part 3 of 6 · All parts →

A football punt, a cricket lofted drive, a javelin throw, water from a fountain — all follow the same curve, and all run on the same handful of formulas. Projectile launched at an angle is projectile physics in its full glory: the general case from which the horizontal launch of Part 2 is just a special case with θ = 0. Part 3 of the Motion in a Plane series builds the complete toolkit — time of flight, maximum height, and range — and then unpacks the two results examiners love most: why 45° wins, and why twin angles land together.

Splitting the Launch: Two Independent Movies

When a ball is kicked at speed u at an angle θ above the horizontal, resolve the launch velocity exactly as we did in Part 2:

• uₓ = u cos θ — the horizontal component. No horizontal force acts (we neglect air resistance), so this component rides flat forever. The ball covers equal horizontal distance in every equal interval of time.
• u_y = u sin θ — the vertical component. Gravity decelerates it at g every second: it climbs, slows, momentarily pauses at the peak, then falls back with ever-increasing speed.

This is the same “two independent movies” trick from Part 2 — the horizontal cruise and the vertical climb-fall story — with one crucial difference: the vertical movie now starts with an upward velocity instead of zero. Every formula below is just the one-dimensional kinematics of Part 1 applied to each movie separately, then stitched back together.

The Full Toolkit: Time of Flight, Maximum Height, Range

These are not magic — here is where each one comes from, so you can rebuild them even under exam pressure:

Time of flight T

The projectile lands back at launch level when its net vertical displacement is zero: 0 = (u sinθ)T − ½gT². One root is T = 0 (the launch moment); the other is T = 2u sinθ/g. Notice T depends only on the vertical component — a harder-thrown ball or a steeper angle means a longer hang.

Maximum height H

At the top, v_y = 0, so 0 = (u sinθ)² − 2gH, giving H = u² sin²θ/2g. This is exactly the “maximum height under upward initial velocity” result from Part 1, with u sinθ playing the role of the launch speed.

Range R

Range is simply the horizontal speed times the flight time: R = uₓT = (u cosθ)(2u sinθ/g). Now use the double-angle identity 2 sinθ cosθ = sin2θ to get the compact classic: R = u² sin2θ/g.

Quick sanity check on the structure of the formulas: T and H contain sinθ (the climbing part only), because they are purely vertical-movie quantities. R contains sin2θ, because range is a compromise between climbing (longer flight time) and pushing forward (horizontal speed). Keep that physical picture in mind and you will never mix the formulas up.

The Trajectory Equation: Why the Path Is a Parabola

Eliminate time between x = (u cosθ)t and y = (u sinθ)t − ½gt², and you get:

y = x tanθ − gx² / (2u²cos²θ)

This is y = (tanθ)x − (constant)x² — the equation of a downward-opening parabola. That single word, “parabola,” is a favourite one-mark answer in JEE Main and NEET. It also explains why the descent mirrors the ascent on level ground: the maths has no preferred direction for time’s arrow within each half of the flight.

The 45° Secret and the Twin Angles

Range R = u² sin2θ/g is biggest when sin2θ is biggest. Since the sine of an angle never exceeds 1, the maximum occurs at 2θ = 90°, i.e. θ = 45°, where R_max = u²/g. Throw a ball at 20 m/s and no angle on Earth will land it farther than 40 m (with g = 10).

And because sin2θ repeats symmetrically about 45°, complementary angles θ and (90° − θ) give identical ranges: sin(2 × 30°) = sin60° and sin(2 × 60°) = sin120° = sin60°. So 30° and 60° give the same range — one flies low and flat, the other high and looping, but both land on the same mark. The 60° version, however, flies higher (H ∝ sin²θ) and hangs longer (T ∝ sinθ). Footballers and javelin throwers instinctively live between these anchors, trading height for distance based on wind, obstacles, and what the situation demands.

What Happens at the Top

At maximum height the vertical movie pauses: v_y = 0. But the horizontal cruise never stopped. So the velocity at the top is u cos θ, directed horizontally — not zero. Meanwhile, the acceleration never took a break either: it is a full g, directed vertically downward, at every instant of the flight, including the peak.

Why does the projectile not “hang” at the top? Because at that instant its velocity is horizontal and gravity bends it downward from there — the peak is a smooth, flat-topped instant on the parabola, not a rest stop. This is precisely the moment where the trajectory is momentarily parallel to the ground.

Solved Examples

This Physics in Your Daily Life

Practice Set (Answers Hidden — Try First)

Frequently Asked Questions

Q. Do T, H, and R depend on the mass of the projectile?
Never. Mass cancels out because gravity gives every mass the same acceleration g. A cricket ball and a cork ball launched identically trace the identical parabola (in vacuum). Air resistance is where mass starts to matter — but JEE/NEET projectile questions ignore it unless stated.

Q. What if the projectile lands below or above the launch level?
The tidy T, H, R formulas assume equal launch and landing heights. If a ball is kicked off a cliff, find T from the vertical equation 0 − h = (u sinθ)T − ½gT² (with h the drop), then compute range as u cosθ × T. That is the Part 2 machinery with a launch angle added.

Q. Is velocity ever zero during flight?
No — not even at the top. Only the vertical component is zero at the peak; the horizontal u cosθ persists the whole flight. Speed is minimum (equal to u cosθ) at the top and maximum (equal to u, by symmetry) at launch and landing.

Q. Why does a real javelin thrown at 45° not match R = u²/g?
Air resistance and the javelin’s aerodynamic shape change the game; real athletes launch near 33–36°. The 45° rule is the ideal, no-air result that all exam questions assume.