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Most projectile questions give you u and θ and ask for range or height. JEE likes to turn this around: here is the path, y = x − x²/20, now find the speed and angle. Students who know only the range and height formulas get stuck. This post shows how to read everything from the trajectory equation, with three worked examples and the slips to avoid.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "Motion in a plane, projectile motion" under kinematics, and the NEET (UG) 2026 syllabus lists projectile motion too. The JEE (Advanced) 2026 brochure has "Kinematics in one and two dimensions (Cartesian coordinates only), projectiles". The equation of the path is Eq. (3.39) in NCERT Class 11 Physics, Motion in a Plane (2026-27 reprint). Check your exam year's syllabus.
Where does the equation come from?
Launch from the origin with speed u at angle θ above the horizontal, ignoring air resistance:
x = (u cos θ) t
y = (u sin θ) t − ½ g t²
Take t = x/(u cos θ) from the first and put it in the second:
y = (tan θ) x − g x² / (2u² cos² θ)
NCERT notes that this is the equation of a parabola, so the path is a parabola.
How do you read u and θ from y = ax − bx²?
Match the two forms term by term:
Angle: tan θ = a.
Horizontal speed: u² cos² θ = g/(2b), so u cos θ = √(g/2b).
Launch speed: since 1/cos² θ = 1 + tan² θ, u² = g(1 + a²)/(2b).
Range: y = 0 again at x = a/b.
Maximum height: at x = a/(2b), y = a²/(4b).
The same equation can be written y = x tan θ (1 − x/R), where R is the range. That factor form is the quickest tool for reverse problems.
Worked example 1: y = x − x²/20
Units are SI and g = 10 m/s².
tan θ = 1, so θ = 45°.
u cos θ = √(10 ÷ (2/20)) = √100 = 10 m/s.
u² = 10 × (1 + 1) ÷ (2/20) = 200, so u = √200 ≈ 14.1 m/s.
Range: 1 ÷ (1/20) = 20 m. Maximum height: 1 ÷ (4/20) = 5 m.
Time of flight: range ÷ horizontal speed = 20 ÷ 10 = 2 s.
Check with the standard results: R = u² sin 2θ/g = 200 × 1 ÷ 10 = 20 m, and T = 2u sin θ/g = 2 × 10 ÷ 10 = 2 s. They agree.
Worked example 2: y = 0.75x − 0.025x²
tan θ = 0.75, so θ ≈ 36.9°, with sin θ = 0.6 and cos θ = 0.8 (a 3-4-5 triangle).
u cos θ = √(10 ÷ 0.05) = √200 ≈ 14.1 m/s.
u² = 200 × (1 + 0.5625) = 312.5, so u ≈ 17.7 m/s.
Range: 0.75 ÷ 0.025 = 30 m. Maximum height: 0.5625 ÷ 0.1 = 5.625 m.
Check: R = 312.5 × sin 2θ ÷ 10 = 312.5 × 0.96 ÷ 10 = 30 m, and H = 312.5 × 0.36 ÷ 20 = 5.625 m.
Worked example 3: a ball that just clears two walls
A ball is launched from the ground and just clears two walls, each 5.0 m high, at x = 10 m and x = 30 m. Find θ, u and the maximum height (g = 10 m/s²).
The path is symmetric, and the two walls are at the same height, so the midpoint between them, x = 20 m, is the top. The range is twice that: R = 40 m.
Use the factor form at x = 10 m: 5 = a × 10 × (1 − 10/40) = 7.5a, so a = tan θ = 2/3 and θ ≈ 33.7°.
b = a/R = (2/3)/40 = 1/60.
u cos θ = √(10 ÷ (2/60)) = √300 ≈ 17.3 m/s, and u² = 300 × (1 + 4/9) ≈ 433.3, so u ≈ 20.8 m/s.
Maximum height: a²/(4b) = (4/9) ÷ (4/60) ≈ 6.67 m.
Check the second wall: y(30) = (2/3) × 30 − 900/60 = 20 − 15 = 5 m.
Which mistakes cost marks?
Using a directly as θ, or a calculator set to radians.
Dropping the cos² θ when solving for u. Using 1 + a² avoids it.
Getting the components wrong. Mikula and Heckler (2014) found that difficulties with trigonometric vector components persist across four populations of introductory physics students, so write u cos θ and u sin θ out each time.
Mismatched g. If the question uses g = 9.8, the numbers change; read the question.
Using the equation for a launch above the ground without moving the origin. It assumes the projectile starts at (0, 0).
A practice routine
Derive the trajectory equation from x(t) and y(t) on a blank page.
Solve y = √3 x − x²/40 (g = 10) for θ, u, R and H, then check with the standard formulas.
Make up a two-wall problem and solve it with the factor form.
Mix in three ordinary projectile questions so you practise choosing a method, not just applying one.
If the shape changes with angle and speed are hard to picture, Fermi's Simulation Lab has interactive simulations for ideas that are easier to see than read. For range and height from u and θ, read our post on projectile range and maximum height, and for slopes, see projectiles on an inclined plane.
Tonight, solve practice question 2 and check that your range matches a/b.
