Projectile motion — horizontal and vertical are independent — Physics

Before this

This leads to

Two balls, one table edge

Put two identical balls at the edge of a table. Flick one so that it shoots off the edge sideways. At the same moment, let the other one simply drop from the edge. Which one hits the floor first?

Most people expect the dropped ball to win. The flicked ball has somewhere to go, so surely it stays in the air longer. Or perhaps it falls faster because it is moving faster.

Neither is true. If air resistance is small, the two balls hit the floor at the same moment. You can hear it: one click, not two. The same would be true for a bullet fired sideways and a bullet dropped from the same height, if there were no air at all. Sideways speed, however large, does not change how long an object takes to fall.

This lecture explains why. The explanation turns every problem in this topic into something you can already do.

What a projectile is

A projectile is an object that has been launched and then moves under gravity alone. Nothing pushes it after launch: no engine, no wind. A kicked football, a stone thrown from a bridge, water from a hose and a long jumper in mid-air are all projectiles, as long as we ignore air resistance. In this topic, we always ignore it.

A projectile's path is a curve. It moves sideways and downwards at the same time. That looks like a new kind of problem, and SUVAT was built for motion in a straight line. So how can SUVAT help?

Two directions, two separate stories

Here is the key idea. Split the motion into two directions: horizontal (sideways) and vertical (up and down). Then ask what acts in each direction.

Horizontally, nothing acts. Gravity pulls straight down, so it has no sideways part. With air resistance ignored, there is no horizontal force at all. So nothing changes the horizontal velocity. The horizontal acceleration is zero, and the horizontal velocity stays the same for the whole flight.

Vertically, gravity acts, exactly as it does on a dropped ball. The vertical acceleration is $g = 9.8 m/s^{2}$ , downwards, whatever the object is doing sideways.

That is it. That is the whole idea. Gravity cannot see the sideways motion, and the sideways motion cannot feel gravity. So the vertical motion of the flicked ball is exactly the vertical motion of the dropped ball. Both start with zero vertical velocity, and both speed up downwards at 9.8 m/s² every second. They are at the same height at every moment, so they land together.

A strobe picture: both balls are shown every 0.25 s until they land, about 2.02 s after leaving a 20 m ledge. The dashed lines join positions at the same moment. At every moment the two balls are at the same height. Only the launched ball also moves sideways, by the same 3 m in every 0.25 s.

Look at the picture in two ways. Read it down, and the gaps between the dots grow: the fall speeds up, as in any drop. Read it across, and the gaps between the blue dots stay equal: the sideways motion is steady. Two different kinds of motion are happening at once, and neither affects the other.

Two SUVAT lists, one clock

In the lecture on solving kinematics problems, the method was to choose a positive direction, list all five SUVAT letters, and then choose an equation. A projectile needs the same method twice: one list for each direction.

The horizontal list: $a = 0$ , so the velocity never changes, and distance = velocity × time.
The vertical list: $a = 9.8 m/s^{2}$ downwards, and the initial vertical velocity depends on the launch. For a horizontal launch, it is zero.

The two lists share only one letter: the time, $t$ . The ball spends the same number of seconds moving sideways as it spends falling, because it is one ball. So the time is the link. Usually you find $t$ from one list and use it in the other.

The vertical list usually gives the time, because it is where the height is. Then the horizontal list turns that time into a distance.

Explore it below. Change the launch speed and step through the flight.

Explore — the same fall at every launch speed

Choose a launch speed. Then step through time. Watch the blue ball's height and the dropped ball's height, and the time when each lands.

Launch speed: 12 m/s

Time: 1.00 s

A ball launched sideways at 12 m/s from a 20 m ledge, beside a dropped ball. Every row is true for both balls, except the horizontal distance, which is zero for the dropped ball.

t (s)	Horizontal distance (m)	Distance fallen (m)	Vertical velocity (m/s, down)
0.5	6.0	1.2	4.9
1.0	12.0	4.9	9.8
1.5	18.0	11.0	14.7
2.02 (lands)	24.2	20.0	19.8

At any launch speed, the distance fallen and the time to land (2.02 s) are the same. Only the horizontal distance changes.

Did you see it? Changing the launch speed stretches the path sideways, but the heights at each moment never change, and the landing time stays at 2.02 s. At 0 m/s, the "launched" ball is simply a second dropped ball.

Stop and think

In cartoons, a character runs off a cliff, keeps going straight for a moment, looks down, and only then falls. Does a real ball rolling off a table do this?

Show the answer

No. Gravity acts from the first instant the ball leaves the table, so the ball starts to fall at once. Its vertical velocity is zero at the edge, so in the first tiny moment it falls very little, which is why the start of the path looks almost flat. But it is already curving down. There is no moment of straight sideways travel. The cartoon gets the horizontal part right (the sideways speed continues) and the vertical part wrong.

Worked examples

In every example, write two lists. For the vertical list, state the positive direction, as in the kinematics lectures. Here, downwards is taken as positive, because everything in the vertical list points down.

Worked Example 1 — A ball kicked off a cliff

A ball is kicked horizontally off a cliff at 12 m/s. The cliff is 20 m high. How far from the foot of the cliff does the ball land?

Vertical list (down positive): $s = 20$ m, $u = 0$ (a horizontal launch has no vertical velocity), $a = 9.8 m/s^{2}$ , $t = ?$ . Use $s = u t + \frac{1}{2} a t^{2}$ :

20 = 0 + 4.9 t^{2} \Rightarrow t^{2} = \frac{20}{4.9} \approx 4.08 \Rightarrow t \approx 2.02 s

Horizontal list: velocity 12 m/s, $a = 0$ , $t \approx 2.02$ s. Distance = $12 \times 2.02 \approx 24.2$ m.

The ball lands about 24 m from the foot of the cliff. Check: 12 m/s for about 2 s is about 24 m.

Worked Example 2 — Working back from the time

A stunt performer runs horizontally off a platform at 5 m/s and is in the air for 1.8 s. How high is the platform, and how far forward does she travel?

This time the time is given, so each list can be solved on its own.

Vertical list (down positive): $u = 0$ , $a = 9.8 m/s^{2}$ , $t = 1.8$ s, $s = ?$

s = \frac{1}{2} \times 9.8 \times 1. 8^{2} = 4.9 \times 3.24 \approx 15.9 m

Horizontal list: distance = $5 \times 1.8 = 9$ m.

The platform is about 15.9 m high, and she lands 9 m forward. Check: 5 m/s for nearly 2 s is nearly 10 m, so 9 m is sensible.

Worked Example 3 — Same table, different speeds

Two marbles roll off a table 0.8 m high, one at 1 m/s and one at 3 m/s. Find each one's time in the air and where it lands.

Vertical list (the same for both): $s = 0.8$ m, $u = 0$ , $a = 9.8 m/s^{2}$ . So $0.8 = 4.9 t^{2}$ , and $t \approx 0.404$ s for both marbles.

Horizontal lists: $1 \times 0.404 \approx 0.40$ m and $3 \times 0.404 \approx 1.21$ m.

Three times the speed gives three times the distance, in exactly the same time. The faster marble does not stay up longer. It simply covers more ground while falling the same 0.8 m.

Worked Example 4 — Where and how fast, partway through

For the ball in Worked Example 1, find its position and its velocity 1.0 s after the kick.

Horizontal: it has moved $12 \times 1.0 = 12$ m sideways, and its horizontal velocity is still 12 m/s.

Vertical (down positive): it has fallen $\frac{1}{2} \times 9.8 \times 1. 0^{2} = 4.9$ m, and its vertical velocity is $v = u + a t = 0 + 9.8 \times 1.0 = 9.8$ m/s, downwards.

So after 1 s, the ball is 12 m out and 4.9 m down, so it is 15.1 m above the ground at the foot of the cliff. It is moving 12 m/s sideways and 9.8 m/s downwards at the same time. Its path is tilted downwards, and it tilts more steeply every second, because the vertical velocity keeps growing while the horizontal velocity stays the same.

The worked examples used correct lists. The working below contains the most common mistake in this topic. Find it.

Spot the error — a stone thrown from a bridge

A stone is thrown horizontally at 15 m/s from a bridge 45 m above a river. How far from the bridge does it land? One line of this working is wrong. Tap it.

The working, with the mistake marked and corrected:

Line 1. Take downwards as positive for the vertical motion. (Correct.)
Line 2. Horizontal list: velocity 15 m/s, a = 0. (Correct.)
Line 3. Vertical list: s = 45 m, u = 15 m/s, a = 9.8 m/s². The mistake: the 15 m/s is horizontal. A horizontal launch has u = 0 in the vertical list.
Line 4. 45 = 15t + 4.9t², so t ≈ 1.86 s. (This follows from the mistake. With u = 0, 45 = 4.9t², so t ≈ 3.03 s.)
Line 5. Distance = 15 × 1.86 ≈ 28.0 m. (It should be 15 × 3.03 ≈ 45.5 m.)

The mistake mixes the two lists. The quick check in the last explanation is worth remembering: a horizontal throw takes exactly as long to land as a drop from the same height.

Three classic mistakes

Trap 1 — Putting the horizontal speed in the vertical list

For a horizontal launch, the initial vertical velocity is zero, however fast the throw. Each velocity belongs in its own list. Only the time is shared.

Trap 2 — Letting gravity slow the sideways motion

Gravity acts straight down, so the horizontal acceleration is zero and the horizontal velocity never changes. Do not write $a = 9.8$ in the horizontal list. (In real life, air resistance does slow the sideways motion a little. Our problems ignore it.)

Trap 3 — "The faster one stays up longer"

The time in the air depends only on the vertical motion: the height and the initial vertical velocity. For horizontal launches from the same height, every speed gives the same time. A faster launch lands further away, not later.

Try it yourself

Work each one on paper first. Take g = 9.8 m/s² and ignore air resistance. Open a hint only when you are stuck.

A ball rolls off a table 1.25 m high at 2 m/s. How long is it in the air, and how far from the table does it land?

Hint 1
Start with the vertical list. The initial vertical velocity is 0.

Hint 2
$1.25 = 4.9 t^{2}$ . Then use the time in the horizontal list.

Full answer
Vertical (down positive): $t^{2} = 1.25 \div 4.9 \approx 0.255$ , so $t \approx 0.505$ s. Horizontal: $2 \times 0.505 \approx 1.01$ m. Check: a little over 1 m in half a second at 2 m/s is sensible.
An aircraft flying horizontally at 60 m/s, 500 m above flat ground, releases a box of supplies. How long does the box take to land, and how far ahead of the release point does it land? Where is the aircraft at that moment?

Hint 1
When the box is released, it has the aircraft's velocity: 60 m/s horizontally, and 0 vertically.

Hint 2
Vertical: $500 = 4.9 t^{2}$ . Then compare the box's horizontal distance with the aircraft's.

Full answer
Vertical: $t^{2} = 500 \div 4.9 \approx 102$ , so $t \approx 10.1$ s. Horizontal: $60 \times 10.1 \approx 606$ m ahead of the release point. The aircraft has also flown at 60 m/s for 10.1 s, so it is directly above the box when the box lands. Seen from the aircraft, the box falls straight down. This is why supplies must be released well before the aircraft reaches the target.
A dart is thrown horizontally at 10 m/s at a board 2.5 m away. The thrower aims straight at the centre. How far below the centre does the dart hit?

Hint 1
This time, the horizontal list gives the time.

Hint 2
$t = 2.5 \div 10 = 0.25$ s. Then find how far the dart falls in that time.

Full answer
Horizontal: $t = 2.5 \div 10 = 0.25$ s. Vertical: fall = $\frac{1}{2} \times 9.8 \times 0.2 5^{2} \approx 0.306$ m. The dart hits about 31 cm below the centre. This is why players aim slightly above where they want to hit. A faster throw spends less time in the air, so it drops less.
The ball in Worked Example 1 (12 m/s off a 20 m cliff) lands after about 2.02 s. How fast is it moving when it lands?

Hint 1
Find the two velocities at landing: horizontal and vertical. The ball is moving with both at once.

Hint 2
The vertical velocity is $9.8 \times 2.02$ . The two velocities are at right angles, so combine them with Pythagoras' theorem.

Full answer
Horizontal velocity: still 12 m/s. Vertical velocity: $v = 0 + 9.8 \times 2.02 \approx 19.8$ m/s downwards (or, without the rounded time, $v^{2} = 2 \times 9.8 \times 20 = 392$ ). The speed is $1 2^{2} + 392 = 536 \approx 23.2$ m/s. Check: the speed must be more than either part (12 and 19.8) and less than their sum (31.8). The answer is.

The bigger idea

The independence of the two directions is not a trick for exams. It follows from a simple fact: a force changes motion only in its own direction. Gravity pulls down, so it changes only the vertical motion. You will see this idea stated properly in the lectures on Newton's laws.

Physicists use it everywhere. To understand any motion in two or three dimensions, they split it into directions at right angles, solve each direction on its own, and join the answers with the shared time. A curved path that looks complicated becomes two straight-line problems. In the next two lectures, the same method finds the range and the greatest height of a projectile, and then handles launches at an angle.

Key concepts in this lesson

A projectile is an object moving under gravity alone after it is launched, with air resistance ignored.
Horizontally there is no force, so the horizontal acceleration is zero and the horizontal velocity stays constant for the whole flight.
Vertically the acceleration is g = 9.8 m/s² downwards, exactly as for an object that is simply dropped.
The two directions are solved as two separate SUVAT lists, and the only quantity they share is the time.
For horizontal launches from the same height, the time in the air is the same at every launch speed, so a faster launch lands further away but not later.