Projectile motion

Two coins, one table

Put one coin at the edge of a table. Flick a second coin so that it flies off the edge sideways, and at the same moment knock the first coin so that it just drops. Listen. The two coins hit the floor together.

The flicked coin travels much farther. Yet it does not stay in the air any longer. Its sideways motion and its fall do not disturb each other. That is the whole secret of projectile motion, and you already met it in the last course.

What a projectile is

A thrown ball, water from a hose, a long jumper in the air: after the launch, the only force on each is gravity. We ignore the air. Such a body is a projectile. We take $g = 9.8\ \text{m/s}^2$, the same along the whole path, and treat the ground as flat.

Split the motion into two parts:

• Horizontally: no force, so no acceleration. The horizontal velocity never changes.
• Vertically: the acceleration is $g$ downwards. This is the free fall you know from the last chapter.

Launched at an angle

The ball leaves the origin at the speed $u$, at the angle $\theta$ above the horizontal. Its two starting parts are

Take upwards as positive, so $a_y = -g$. After a time $t$:

Now watch it fly. The simulation also lets you launch from a height, which we come back to below.

Three numbers for every flight

These hold when the ball lands at the same level it was thrown from.

Time of flight. It lands when $y = 0$ again: $(u\sin\theta)\,t - \tfrac{1}{2}gt^2 = 0$. So

The rise takes half of this, $\dfrac{u\sin\theta}{g}$, and the fall takes the other half.

Greatest height. At the top, $v_y = 0$. From $v_y^2 = u_y^2 - 2gy$:

Range. The horizontal distance covered in the time $T$:

Since $2\sin\theta\cos\theta = \sin 2\theta$, this is

Worked example. A ball is thrown at $24\ \text{m/s}$, at $50^\circ$ above level ground.

1. Split: $u_x = 24\ \text{m/s}\cos 50^\circ = 15.43\ \text{m/s}$ and $u_y = 24\ \text{m/s}\sin 50^\circ = 18.39\ \text{m/s}$.
2. Time: $T = \dfrac{2 \times 18.39\ \text{m/s}}{9.8\ \text{m/s}^2} = 3.75\ \text{s}$.
3. Height: $H = \dfrac{(18.39\ \text{m/s})^2}{2 \times 9.8\ \text{m/s}^2} = 17.3\ \text{m}$.
4. Range: $R = 15.43\ \text{m/s} \times 3.75\ \text{s} = 57.9\ \text{m}$.
5. Check with the other formula: $\dfrac{(24\ \text{m/s})^2\sin 100^\circ}{9.8\ \text{m/s}^2} = 57.9\ \text{m}$. They agree.

The path is a parabola

From $x = (u\cos\theta)\,t$ we get $t = \dfrac{x}{u\cos\theta}$. Put it into the equation for $y$:

This has the form $y = ax - bx^2$: a parabola, opening downwards.

Thrown sideways from a height

A stone is thrown horizontally at $u$ from a cliff of height $h$. Now $u_x = u$ and $u_y = 0$. Vertically it is simply dropped, so the time to land depends only on the height:

It lands $x = u\,t$ from the foot of the cliff. Just before landing, $v_x = u$ and $v_y = gt$ downwards.

Worked example. A stone is thrown horizontally at $12\ \text{m/s}$ from a cliff $30\ \text{m}$ high.

1. Time: $t = \sqrt{\dfrac{2 \times 30\ \text{m}}{9.8\ \text{m/s}^2}} = 2.47\ \text{s}$.
2. Distance: $x = 12\ \text{m/s} \times 2.47\ \text{s} = 29.7\ \text{m}$.
3. Landing velocity: $v_y = 9.8\ \text{m/s}^2 \times 2.47\ \text{s} = 24.2\ \text{m/s}$. So $v = \sqrt{(12\ \text{m/s})^2 + (24.2\ \text{m/s})^2} = 27.1\ \text{m/s}$, at $\tan^{-1}\dfrac{24.2\ \text{m/s}}{12\ \text{m/s}} = 63.7^\circ$ below the horizontal.

Watch out. These traps appear in almost every exam.

• At the top, the velocity is not zero. It is horizontal, equal to $u\cos\theta$.
• At the top, the acceleration is still $g$, downwards. Gravity never switches off.
• $T = \dfrac{2u\sin\theta}{g}$ only works when the ball lands at its launch level. From a cliff or onto a roof, solve $y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2$ for the real landing height. The next lesson shows how.

Practise with new numbers

Try it in Python

The program works out the worked example, and draws the path.

Remember these

• Horizontal: $v_x = u\cos\theta$, steady. Vertical: free fall, $a_y = -g$.
• $T = \dfrac{2u\sin\theta}{g}$, $H = \dfrac{u^2\sin^2\theta}{2g}$, $R = \dfrac{u^2\sin 2\theta}{g}$, on level ground.
• Thrown horizontally from $h$: $t = \sqrt{2h/g}$, whatever the throwing speed.
• The path is a parabola.

Check yourself

Try each one on your own first. If you are stuck, open a hint. When you have an answer, check it with Show the answer.

1. At the top of its path, what are a projectile's velocity and acceleration?

2. Two balls are thrown horizontally from the same height, one at $5\ \text{m/s}$ and one at $15\ \text{m/s}$. Which lands first?

3. A ball is thrown at $15\ \text{m/s}$ at $53^\circ$ above level ground. Taking $\cos 53^\circ = 0.6$ and $\sin 53^\circ = 0.8$, find $u_x$, $u_y$ and the time of flight.