Anything thrown, kicked or fired at an angle follows the same path, and the trick to analysing it is to stop treating it as one motion. Split it into a horizontal motion with no acceleration and a vertical motion under gravity, and every question becomes two easy questions that share a clock.
The idea the whole topic rests on
The horizontal and vertical motions are completely independent. Gravity pulls downwards, so it changes the vertical velocity and does nothing at all to the horizontal one. That means:
Horizontally: the velocity u cos θ never changes. Distance is simply speed × time.
Vertically: the object starts at u sin θ and decelerates at g, stops rising, then falls back. This is ordinary constant-acceleration motion, and the equations of motion apply to it directly.
The only thing the two motions share is time. That shared clock is what links them, and almost every projectile question is solved by finding a time from one direction and using it in the other.
A consequence students often find surprising: a ball fired horizontally from a table and a ball simply dropped from the same height hit the floor at the same moment. The horizontal motion has no say in how long the fall takes.
The three standard results
| Quantity | Formula | Where it comes from |
|---|---|---|
| Time of flight, T | 2u sin θ ⁄ g | Vertical displacement returns to zero |
| Greatest height, H | u² sin²θ ⁄ 2g | Vertical velocity is zero at the top |
| Range, R | u² sin 2θ ⁄ g | Horizontal speed × time of flight |
All three assume the projectile lands at the same height it was launched from. If it is thrown off a cliff or lands on a roof, they do not apply and you must go back to the equations of motion applied separately to each direction.
Two facts fall straight out of the range formula and are worth carrying into an exam. Since sin 2θ is largest when 2θ = 90°, the greatest range comes at θ = 45°. And because sin 2θ has the same value for θ and for 90° − θ, two different angles give the same range — 30° and 60° land in the same place, the steeper one simply taking longer to get there.
A worked example
A ball is kicked at 20 m/s at 35° above the horizontal, from ground level. Take g = 9.81 m/s².
Components first.
Horizontal: uₓ = 20 cos 35° = 20 × 0.8192 = 16.38 m/s
Vertical: u_y = 20 sin 35° = 20 × 0.5736 = 11.47 m/s
Time of flight. It rises until the vertical velocity reaches zero, which takes u_y ⁄ g = 11.47 ⁄ 9.81 = 1.169 s, then takes the same again to fall back:
T = 2 × 1.169 = 2.34 s
Greatest height. Using v² = u² + 2as vertically, with v = 0 at the top:
H = u_y² ⁄ 2g = 11.47² ⁄ (2 × 9.81) = 131.6 ⁄ 19.62 = 6.71 m
Range. Horizontal speed is constant, so it is just speed × time:
R = 16.38 × 2.34 = 38.3 m
Check it against the direct formula: R = u² sin 2θ ⁄ g = 400 × sin 70° ⁄ 9.81 = 400 × 0.9397 ⁄ 9.81 = 38.3 m ✓
Where marks get lost
Using sin where cos belongs. The vertical component takes sin, the horizontal takes cos, because the angle is measured from the horizontal. If a question measures the angle from the vertical instead — and some do — they swap over. Read the diagram.
Calculator in radians. sin 35 in radian mode is −0.428 rather than 0.574. Every answer that follows is wrong and none of them look obviously wrong. Check the mode before you start.
Using the standard formulas when the landing height differs. Off a cliff, onto a wall, into a basket — none of the three results above apply. Go back to first principles with the equations of motion.
Forgetting the second angle. If a question asks for the angle to reach a given range, there are usually two valid answers, θ and 90° − θ. Giving only one is giving half the answer.
Thinking acceleration is zero at the top. The vertical velocity is zero there; the acceleration is 9.81 m/s² downwards for the whole flight, which is exactly why the object does not stay up there.
Questions students actually ask
Why does 45° give the maximum range?
Because range depends on sin 2θ, and sine peaks at 90°. That needs 2θ = 90°, so θ = 45°. It balances staying in the air long enough against keeping enough horizontal speed — steeper gives more time but less speed, shallower the reverse.
Why do two different angles give the same range?
Because sin 2θ takes the same value at θ and at 90° − θ. So 30° and 60° produce the same range, as do 20° and 70°. The steeper launch simply spends longer in the air on a higher path.
Does the mass of the projectile matter?
Not once air resistance is ignored. Gravity accelerates everything at the same rate, so a cannonball and a tennis ball launched identically follow the same path. Mass starts to matter as soon as drag is included, which is why a real table-tennis ball behaves nothing like this.
What happens to the velocity at the highest point?
The vertical component is zero, but the horizontal component is unchanged. So the projectile is still moving at u cos θ at the top — it is at its slowest there, not stationary.
How do I handle a projectile launched from a height?
Do not use the standard range and time formulas — they assume it lands where it started. Apply the equations of motion to the vertical direction with the correct displacement, find the time, then multiply by the horizontal velocity.
Related formulas
The equations of motion · Newton’s second law · Pythagoras’ theorem
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