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Projectile questions on flat ground are among the most predictable in Class 11 physics and JEE Main, yet students still lose marks on them. The usual causes are a formula used outside the case it was derived for, a sine where a cosine belongs, or the belief that the ball stops at the top. This post derives the three standard results, works one launch at two angles with full numbers, and lists the traps.
Where does this sit in the syllabus?
The JEE (Main) 2026 syllabus lists projectile motion under Unit 2, Kinematics, and the NCERT Class 11 chapter on motion in a plane derives time of flight, maximum height and range in Section 3.9. Check the syllabus for your own exam year.
Why can you treat the two directions separately?
Ignoring air resistance, the only acceleration is g, straight down. OpenStax University Physics puts the key idea plainly: motions along perpendicular axes are independent and can be analysed separately. So a projectile is two simpler motions running at once:
Horizontal: no acceleration, so u_x = u cos θ stays constant.
Vertical: acceleration −g, starting at u_y = u sin θ, exactly like a ball thrown straight up.
Time is the only thing the two motions share.
How are the three formulas derived?
Time to the top: the vertical velocity u sin θ − gt becomes zero at t = u sin θ/g.
Time of flight: on level ground, the way down mirrors the way up, so T = 2u sin θ/g.
Maximum height: from v² = u² − 2gH with v_y = 0 at the top, H = u² sin² θ/2g.
Range: horizontal speed × time of flight = u cos θ × 2u sin θ/g = u² sin 2θ/g, using 2 sin θ cos θ = sin 2θ.
Since sin 2θ is largest at 90°, the range is greatest at θ = 45°, where R_max = u²/g. NCERT and OpenStax both show this. OpenStax adds two caveats: the range formula is valid only for launch and landing at the same height, and with air resistance the angle for maximum range is somewhat smaller than 45°.
Worked example: the same throw at 30° and 60°
A ball is launched at 20 m/s from level ground. Take g = 10 m/s², sin 30° = 0.5 and cos 30° = sin 60° ≈ 0.866.
At 30°:
Components: u_x = 20 × 0.866 = 17.32 m/s, u_y = 20 × 0.5 = 10 m/s.
Time of flight: T = 2 × 10/10 = 2 s.
Maximum height: H = 10²/(2 × 10) = 5 m.
Range: R = 20² × sin 60°/10 = 40 × 0.866 ≈ 34.6 m.
Check: u_x × T = 17.32 × 2 = 34.64 m. The two methods agree.
At 60°:
Components: u_x = 10 m/s, u_y = 17.32 m/s.
Time of flight: T = 2 × 17.32/10 ≈ 3.46 s.
Maximum height: H = 17.32²/20 = 300/20 = 15 m.
Range: R = 40 × sin 120° = 40 × 0.866 ≈ 34.6 m.
The ranges are equal because sin 120° = sin 60°. Any two angles that add to 90° give the same range, as NCERT shows for angles above and below 45° by equal amounts. The heights are not equal: the 60° throw goes three times as high, since H depends on sin² θ and (0.866/0.5)² = 3.
At 45°: R = u²/g = 400/10 = 40 m, and H = u²/4g = 10 m. That is the longest range for this launch speed.
How fast is the ball moving at the top?
Not zero. At the top only the vertical velocity is zero. For the 30° throw, the speed at the top is u_x = 17.32 m/s.
At other moments, combine the two components. At t = 1.5 s in the 30° throw, v_y = 10 − 10 × 1.5 = −5 m/s (moving down) and v_x = 17.32 m/s, so the speed is √(300 + 25) = √325 ≈ 18.0 m/s.
Two relations worth checking
tan θ = 4H/R. For the 30° throw: 4 × 5/34.64 ≈ 0.577 = tan 30°.
R² = 16 H₁H₂ for a pair of complementary angles: 16 × 5 × 15 = 1200, and 34.64² ≈ 1200.
Derive both yourself from the three formulas rather than memorising them. They make quick checks in multiple-choice questions.
Common mistakes and how to avoid them
Saying the velocity is zero at the top. Only the vertical component is.
Swapping sine and cosine. The horizontal component uses cos θ when θ is measured from the horizontal. If the angle is given from the vertical, swap them.
Using the range formula off level ground. For a slope, see our post on projectiles on an inclined plane.
Calculator in radians. Check the mode before the first sine.
Doubling the time to the top when landing height differs. Only valid on level ground.
A practice routine
Day 1: Derive T, H and R from scratch on a blank page without notes.
Day 2: Recompute the worked example at 15° and 75°. Confirm the ranges match.
Day 3: Five problems asking for velocity at a given time or height.
Day 4: Three problems where the landing point is higher or lower than the launch point.
Predictions are easier to trust once you have watched the trajectory change. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, and our post on learning projectile motion with a simulation gives an activity with predictions to test.
Before your next session, write down the range at 15° for a 20 m/s launch, then check it is 20 m.
