Projectile Motion — Range, Height and Time of Flight
Projectile motion has a reputation for being hard, and it is undeserved. It is not a new chapter of physics at all — it is straight-line motion, done twice, in two directions that never speak to each other. Once you accept that the horizontal and vertical motions are completely independent, all three standard formulas fall out of the kinematics equations you already know, and you can rebuild them in the exam hall if your memory fails.
The one idea everything rests on
Gravity pulls straight down. It has no sideways part. So:
- Horizontally there is no acceleration. The horizontal velocity u cosθ stays exactly the same for the whole flight, and horizontal distance is simply speed × time.
- Vertically the acceleration is −g throughout, and the SUVAT equations apply just as they would to a ball thrown straight up.
- Time is the one thing they share. That is the bridge: find the time from the vertical motion, then use it in the horizontal motion.
This is why a bullet fired horizontally and a bullet simply dropped from the same height hit the ground at the same moment. The forward speed changes where it lands, never when.
The formulas (for launch and landing at the same height)
Time of flight: T = 2u sinθ / g
Maximum height: H = u² sin²θ / (2g)
Range: R = u² sin 2θ / g
Trajectory: y = x tanθ − g x² / (2u² cos²θ) (a parabola)
| Symbol | Meaning | SI unit |
|---|---|---|
| u | Launch speed | m/s |
| θ | Launch angle, measured from the horizontal | degrees |
| g | Acceleration due to gravity | m/s² |
| T | Total time in the air | s |
| H | Maximum height above the launch point | m |
| R | Horizontal range | m |
A note on g. Most CBSE, ICSE and NEET papers use g = 9.8 m/s²; many JEE and some textbook problems use g = 10 m/s² to keep the arithmetic clean. The two give answers that differ by about 2%. Use whichever value your question states and write down which one you used. Every worked example below uses g = 9.8 m/s².
Worked example 1 — the standard question
Question: A ball is thrown at 20 m/s at 30° to the horizontal. Find the time of flight, the maximum height and the range.
Components first:
ux = 20 cos 30° = 20 × 0.86603 = 17.321 m/s
uy = 20 sin 30° = 20 × 0.5 = 10 m/s
Time of flight: T = 2uy / g = (2 × 10) ÷ 9.8 = 20 ÷ 9.8 = 2.041 s
Maximum height: H = uy² / (2g) = 100 ÷ (2 × 9.8) = 100 ÷ 19.6 = 5.102 m
Range, route A (horizontal speed × time):
R = 17.321 × 2.041 = 35.35 m
Route B (the standard formula):
R = u² sin 2θ / g = (400 × sin 60°) ÷ 9.8 =
(400 × 0.86603) ÷ 9.8 = 346.41 ÷ 9.8 = 35.35 m. ✔
Note that sin 2θ means the sine of twice the angle — sin 60°, not 2 sin 30°. Those are 0.866 and 1.000, and confusing them is a guaranteed lost mark.
Worked example 2 — why 45° gives the longest range
Question: The same ball, same 20 m/s, now launched at 45°. Find T, H and R, and explain why this angle is special.
ux = uy = 20 × 0.70711 = 14.142 m/s
T = (2 × 14.142) ÷ 9.8 = 28.284 ÷ 9.8 = 2.886 s
H = 14.142² ÷ 19.6 = 200 ÷ 19.6 = 10.20 m
R = u² sin 90° / g = 400 × 1 ÷ 9.8 = 40.82 m
Check: R = uxT = 14.142 × 2.886 = 40.82 m. ✔
Why 45°? Range depends on sin 2θ, and the sine of anything never exceeds 1. It reaches exactly 1 when 2θ = 90°, i.e. θ = 45°. So 45° is the maximum-range angle, and Rmax = u²/g — on level ground, with air resistance ignored. Real throws and real projectiles peak a little below 45° because of drag.
Worked example 3 — complementary angles
Question: Show that 20 m/s at 60° gives the same range as 20 m/s at 30°, and compare the two flights.
Range at 60°: R = 400 × sin 120° ÷ 9.8. Since sin 120° = sin 60° = 0.86603, R = 346.41 ÷ 9.8 = 35.35 m — identical to example 1. ✔
This works for any pair adding to 90°, because sin 2θ and sin(180° − 2θ) are equal. 20° and 70° give the same range; so do 15° and 75°.
But the flights are nothing alike:
| θ = 30° | θ = 60° | |
|---|---|---|
| uy | 10 m/s | 17.321 m/s |
| Time of flight | 2.041 s | 3.535 s |
| Maximum height | 5.102 m | 15.31 m |
| Range | 35.35 m | 35.35 m |
Check on the 60° row: T = (2 × 17.321) ÷ 9.8 = 34.641 ÷ 9.8 = 3.535 s; H = 17.321² ÷ 19.6 = 300 ÷ 19.6 = 15.31 m; R = uxT = 10 × 3.535 = 35.35 m. ✔
The steep throw goes three times as high and stays up 73% longer, and still lands in the same place. That is a favourite examiner trap, because it looks impossible until you separate the two directions.
Worked example 4 — horizontal projection from a height
Question: A ball is thrown horizontally at 15 m/s from the top of a 45 m cliff. Find the time to land, the horizontal distance travelled, and the speed and direction on landing.
Here uy = 0 at launch, so the standard T and R formulas do not apply — launch and landing are at different heights. Go back to first principles.
Time (vertical motion): h = ½gt²
45 = ½(9.8)t² = 4.9 t²
t² = 45 ÷ 4.9 = 9.1837 → t = 3.031 s
Horizontal distance: x = uxt = 15 × 3.031 = 45.46 m
Landing velocity components:
vx = 15 m/s (unchanged all through the flight)
vy = gt = 9.8 × 3.031 = 29.70 m/s downward
Speed: v = √(15² + 29.70²) = √(225 + 882.0) = √1107.0 = 33.27 m/s
Check by energy conservation: v² = u² + 2gh = 225 + 2(9.8)(45) = 225 + 882 = 1107, so v = 33.27 m/s. ✔ An entirely different chapter, the same number.
Direction: tanα = 29.70 ÷ 15 = 1.980, so α = 63.2° below the horizontal.
Common mistakes that cost marks
- Applying R = u²sin2θ/g when the landing point is lower than the launch point. All three boxed formulas assume equal heights. A cliff, a table or a tower question must be solved from components, as in example 4.
- Reading sin 2θ as 2 sinθ. They are different numbers for every angle except 0°.
- Thinking the horizontal velocity decreases. With no air resistance it is constant for the entire flight. Only vy changes.
- Saying the velocity at the top is zero. Only the vertical component is zero there. The ball is still moving forward at u cosθ, so its speed at the peak is u cosθ, not 0.
- Using time of flight where time to maximum height is needed. The peak comes at T/2 for a symmetric flight, not T.
- Calculator in radian mode. sin 30 in radians is −0.988, not 0.5. Check the mode before the first trigonometric key-press.
- Mixing g = 9.8 and g = 10 inside one solution. Pick one, state it, keep it.
Where projectile motion appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Motion in a plane — derivations of T, H and R, plus the parabolic trajectory equation |
| NEET | Single-step questions on range ratios, complementary angles and maximum height |
| JEE Main & Advanced | Projection on an inclined plane, relative motion of two projectiles, questions with air-drag reasoning |
| GATE (engineering papers) | Particle dynamics and trajectory problems built on the same decomposition |
| Class 12 board practicals | Verifying the parabolic path and estimating g from measured range and angle |
The exam habit worth building
Before writing a single formula, draw the launch point, mark ux and uy, and write "vertical: a = −g" and "horizontal: a = 0" beside them. Those two lines take ten seconds and turn every projectile question into two easy SUVAT problems joined by a shared time. Students who skip that step are the ones who reach for R = u²sin2θ/g in a cliff question and lose the whole answer.
Check your working in seconds. Projectile motion is kinematics, applied separately to each direction — so use the Kinematics tool twice. Feed it uy, a = −9.8 and s = 0 to get the time of flight, then ux, a = 0 and that time to get the range.
Open the Kinematics Calculator →Projectile motion is where a shaky grip on sin, cos and resolution of vectors finally shows up as lost marks. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre and online classes across India — details at abcchemistry.in.