Mathematics/Lessons/Projectiles/Horizontal Projection
Mechanics · Projectiles · Year 2

Horizontal Projection

A particle launched sideways from a height doesn't fall straight down — gravity pulls it downward while it keeps drifting horizontally at a constant speed. Treating these two motions separately is the key to the whole topic.

9 min read 2 worked examples

Splitting the motion into two independent parts

Imagine a ball rolling off the edge of a table. The instant it leaves the table, two things happen at once: gravity starts pulling it downward, and it keeps travelling horizontally at whatever speed it had when it left the edge. The remarkable thing — and the whole basis of this topic — is that these two motions don't affect each other.

We model the object as a particle (so we can ignore its size and shape) moving freely under gravity, with air resistance ignored. Throughout, take $g=9.8\ \text{m s}^{-2}$ unless told otherwise.

The two components of horizontal projection

Horizontal: no force acts horizontally, so the horizontal velocity stays constant. Horizontal distance is just $x = Ut$, where $U$ is the (constant) horizontal speed.

Vertical: the particle starts with zero vertical velocity and accelerates downward at $g$. This is an ordinary SUVAT problem with $u=0$ and $a=g$.

Because the two directions are independent, you almost always work through a horizontal-projection question in the same order: use the vertical equation to find the time of flight $t$ (since that's the direction with the acceleration), then substitute $t$ into the horizontal equation to find distance.

Worked example 1

Time of flight and horizontal distance
A particle is projected horizontally with speed $18\ \text{m s}^{-1}$ from a point $44.1\ \text{m}$ above a horizontal plane. Find (a) the time the particle takes to reach the plane, and (b) the horizontal distance it travels before landing.
MethodVertical motion first, then horizontal
1
Take downward as positive. Vertically: $u=0$, $a=9.8$, $s=44.1$.
The particle starts with no vertical velocity — only the horizontal launch speed.
2
Using $s=ut+\tfrac12 at^2$: $44.1 = 0+\tfrac12(9.8)t^2 = 4.9t^2$.
This is the equation that actually contains $t$, so solve it first.
3
$t^2 = \dfrac{44.1}{4.9} = 9 \Rightarrow t = 3\ \text{s}$.
Take the positive root — time can't be negative.
4
Horizontally, the speed is constant at $18\ \text{m s}^{-1}$, so $x = Ut = 18(3) = 54\ \text{m}$.
Now that $t$ is known, the horizontal distance follows immediately — gravity plays no part in this calculation.
▪ solved
A particle is launched horizontally with velocity U from a point O at height h above a horizontal plane. A curved trajectory shows it falling and landing at point A, a horizontal distance x from directly below O.
The particle leaves O horizontally with speed U. It falls a height h while travelling a horizontal distance x before landing at A.

Distance from the point of projection

Sometimes you're asked how far the particle is from where it started — not just how far it's travelled sideways, or how far it's fallen, but the straight-line distance. Since the horizontal and vertical displacements are perpendicular to each other, you find this using Pythagoras' theorem on the horizontal and vertical distances travelled up to that moment.

Worked example 2

Displacement after a given time
A particle is projected horizontally with speed $12\ \text{m s}^{-1}$. Find, after $2\ \text{s}$: (a) the horizontal and vertical distances travelled, and (b) the particle's distance from the point of projection, to 2 significant figures.
MethodFind each component, then combine with Pythagoras
1
Horizontal distance: $x = Ut = 12(2) = 24\ \text{m}$.
Constant horizontal speed — no need for a SUVAT equation.
2
Vertical distance: $u=0$, $a=9.8$, $t=2$, so $y=\tfrac12(9.8)(2)^2 = 19.6\ \text{m}$.
Only gravity acts vertically, starting from rest in that direction.
3
The horizontal and vertical displacements are perpendicular, so the distance $d$ from $O$ satisfies $d^2 = x^2+y^2 = 24^2+19.6^2$.
This is not the same as the distance fallen or the distance travelled sideways — it's the direct line back to the start.
4
$d^2 = 576+384.16 = 960.16 \Rightarrow d = 31\ \text{m}$ (2 s.f.).
Take the square root and round as instructed.
▪ solved
Distance ≠ distance fallen

Notice $d=31\ \text{m}$ is bigger than either $x=24\ \text{m}$ or $y=19.6\ \text{m}$ on their own — that's exactly what Pythagoras guarantees for a right-angled triangle, and it's a quick way to check an answer looks sensible.

Common mistake

Using the launch speed $U$ as the initial vertical speed. In horizontal projection, $U$ only ever appears in the horizontal equation $x=Ut$ — the vertical motion always starts from $u=0$, because the particle has no vertical velocity at the instant it's launched.

Check yourself

A particle is projected horizontally with speed $U\ \text{m s}^{-1}$ from a point $19.6\ \text{m}$ above a horizontal plane. It lands at a horizontal distance of $60\ \text{m}$ from the point directly below the point of projection. Find the value of $U$ and the time of flight.

Vertically: $19.6 = \tfrac12(9.8)t^2 = 4.9t^2 \Rightarrow t^2=4 \Rightarrow t=2\ \text{s}$.

Horizontally: $60 = U(2) \Rightarrow U = 30\ \text{m s}^{-1}$.

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