Year 2 Mechanics · Chapter 3

Projectiles

Specification: Edexcel Paper 3 §7.5 — Projectiles · AQA section Q5

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In this chapter

1. Projected horizontally
2. Projected at an angle
3. The general formulae
4. The path
5. Launched above the ground
Practice

Key facts — this chapter

Treat the horizontal and vertical motion separately. Ignoring air resistance, the only acceleration is $g$ vertically downwards.
Projected at speed $U$ at angle $\alpha$ above the horizontal: horizontal component $U\cos\alpha$ , which never changes; vertical component $U\sin\alpha$ , which changes at $g$ .
The time links the two directions: find it from one, use it in the other.
On level ground: time of flight $T = \dfrac{2U\sin\alpha}{g}$ , greatest height $H = \dfrac{U^2\sin^2\alpha}{2g}$ , range $R = \dfrac{U^2\sin 2\alpha}{g}$ , greatest at $\alpha = 45^\circ$ .
The path is a parabola: $y = x\tan\alpha - \dfrac{gx^2}{2U^2\cos^2\alpha}$ .
The velocity at any time has components $U\cos\alpha$ and $U\sin\alpha - gt$ : use Pythagoras for the speed and $\tan^{-1}$ for the direction.

A thrown ball moves forwards and up and down at the same time. The trick is that these two motions don’t interfere: horizontally nothing acts on the ball, so it keeps a steady speed, while vertically it rises and falls under gravity exactly as in Year 1. Solve each direction on its own, joined by the time.

Modelling assumption

The ball is modelled as a particle with no air resistance, moving under a constant $g = 9.8\text{ m s}^{-2}$ . Real balls spin and are slowed by the air, so they fall short of these predictions, especially over long distances.

1. Projected horizontally

A ball thrown horizontally at 12 metres per second from the top of a 20 metre cliff, curving down to land on the ground 24.2 metres from the foot of the cliff. Dots mark its position every quarter of a second, evenly spaced across but further apart as it falls — Example 3.1: evenly spaced across, speeding up downwards. The dots are a quarter of a second apart.

Worked example 3.1 · Thrown from a cliff

A ball is thrown horizontally at $12\text{ m s}^{-1}$ from the top of a vertical cliff $20$ m high. Find how long it takes to reach the ground, how far from the foot of the cliff it lands, and its speed as it lands.

Vertically, it starts with no vertical velocity and falls $20$ m:
$\begin{aligned} 20 &= \tfrac{1}{2} \times 9.8\,t^2 \\ t &= 2.02 \text{ s (3 s.f.)} \end{aligned}$
Horizontally, it moves at a steady $12\text{ m s}^{-1}$ : $x = 12 \times 2.020\ldots = 24.2$ m.
Landing: the horizontal velocity is still $12$ , and the vertical velocity is $9.8 \times 2.020\ldots = 19.79\ldots$ , so the speed is $\sqrt{12^2 + 19.79\ldots^2} = 23.2\text{ m s}^{-1}$ (3 s.f.).

In context

The ball lands $24.2$ m from the cliff after $2.02$ s, moving at $23.2\text{ m s}^{-1}$ at $58.8^\circ$ below the horizontal.

2. Projected at an angle

Components of the velocity

A particle projected at speed $U$ at an angle $\alpha$ above the horizontal starts with

\begin{aligned} &U\cos\alpha \text{ horizontally (constant)} \\ &U\sin\alpha \text{ vertically (acceleration } g \text{ down)} \end{aligned}

A parabolic path from the ground at 25 metres per second and 40 degrees, rising to a greatest height of 13.2 metres halfway along and landing 62.8 metres away. The initial velocity is shown with its horizontal and vertical components — Example 3.2: the greatest height comes halfway, and the landing mirrors the launch.

Worked example 3.2 · Time, range and height

A ball is kicked from level ground at $25\text{ m s}^{-1}$ at $40^\circ$ above the horizontal. Find its time of flight, its range and its greatest height.

Vertically, with up positive, $u = 25\sin 40^\circ$ and $a = -9.8$ . It lands when the vertical displacement is zero again:
$\begin{aligned} 0 &= 25\sin 40^\circ\,t - 4.9t^2 \\ t &= \frac{25\sin 40^\circ}{4.9} = 3.28 \text{ s (3 s.f.)} \end{aligned}$
Horizontally: range $= 25\cos 40^\circ \times 3.279\ldots = 62.8$ m.
At the top the vertical velocity is zero. From $v^2 = u^2 + 2as$ :
$\begin{aligned} 0 &= (25\sin 40^\circ)^2 - 19.6H \\ H &= 13.2 \text{ m (3 s.f.)} \end{aligned}$

Common mistake

At the top of the flight the vertical velocity is zero, but the ball is still moving: its horizontal velocity, $U\cos\alpha$ , hasn’t changed. The speed at the top is $U\cos\alpha$ , not zero.

Worked example 3.3 · The velocity after one second

For the ball in Example 3.2, find its speed and direction of motion $1$ s after it is kicked.

\begin{aligned} v_x &= 25\cos 40^\circ = 19.15\ldots \\ v_y &= 25\sin 40^\circ - 9.8 = 6.26\ldots \end{aligned}

The speed is $\sqrt{v_x^2 + v_y^2} = 20.2\text{ m s}^{-1}$ , at $\tan^{-1}\tfrac{v_y}{v_x} = 18.1^\circ$ above the horizontal: $v_y$ is still positive, so the ball is still rising.

Launch a ball from level ground. Set its speed and angle: the path is drawn to scale, with a dot every half second, and the view fits the longest throw at that speed. The dashed path is the launch at the complementary angle, 90° minus yours, which lands in exactly the same place.

3. The general formulae

Doing Example 3.2 with letters instead of numbers gives formulae for any launch from level ground. You may be asked to derive them, so know the steps, not just the results.

Launch from level ground

For speed $U$ at angle $\alpha$ to the horizontal:

\begin{aligned} T &= \frac{2U\sin\alpha}{g} \\ H &= \frac{U^2\sin^2\alpha}{2g} \\ R &= \frac{U^2\sin 2\alpha}{g} \end{aligned}

Worked example 3.4 · Deriving the range

Show that the range on level ground is $\dfrac{U^2\sin 2\alpha}{g}$ , and deduce the angle that gives the greatest range.

Vertically, $0 = U\sin\alpha\,T - \tfrac{1}{2}gT^2$ , so $T = \dfrac{2U\sin\alpha}{g}$ (the other root, $T = 0$ , is the launch).
Horizontally,
$\begin{aligned} R &= U\cos\alpha \times \frac{2U\sin\alpha}{g} \\ &= \frac{U^2 \times 2\sin\alpha\cos\alpha}{g} \\ &= \frac{U^2\sin 2\alpha}{g} \end{aligned}$
using $\sin 2\alpha = 2\sin\alpha\cos\alpha$ from Chapter 10 of Year 2 Pure.
$\sin 2\alpha$ is greatest, $1$ , when $2\alpha = 90^\circ$ . So the greatest range is $\dfrac{U^2}{g}$ , at $\alpha = 45^\circ$ .

Because $\sin 2\alpha = \sin(180^\circ - 2\alpha)$ , the angles $\alpha$ and $90^\circ - \alpha$ give the same range: a high lob and a low drive can land in the same place.

4. The path

The horizontal and vertical positions are both functions of $t$ : a pair of parametric equations, as in Chapter 11 of Year 2 Pure. Eliminating $t$ gives the Cartesian equation of the path.

The equation of the path

y = x\tan\alpha - \frac{gx^2}{2U^2\cos^2\alpha}

a parabola, taking the launch point as the origin.

Why it works

$x = U\cos\alpha\,t$ gives $t = \dfrac{x}{U\cos\alpha}$ . Substitute into $y = U\sin\alpha\,t - \tfrac{1}{2}gt^2$ : the first term becomes $x\tan\alpha$ and the second $\dfrac{gx^2}{2U^2\cos^2\alpha}$ .

Worked example 3.5 · Over the wall?

A ball is thrown from ground level at $14\text{ m s}^{-1}$ at an angle $\alpha$ above the horizontal, where $\tan\alpha = \tfrac{3}{4}$ . Find the equation of its path, and decide whether it clears a wall $2$ m high, $10$ m away.

$\cos\alpha = \tfrac{4}{5}$ , so $2U^2\cos^2\alpha = 2 \times 196 \times \tfrac{16}{25} = 250.88$ , and
$y = \tfrac{3}{4}x - \frac{9.8x^2}{250.88} = \tfrac{3}{4}x - \tfrac{5}{128}x^2$
At $x = 10$ : $y = 7.5 - 3.906\ldots = 3.59$ m (3 s.f.).

In context

The ball is $3.59$ m high as it reaches the wall, so it clears the $2$ m wall.

5. Launched above the ground

When the launch point is above the landing point, the flight doesn’t end when the vertical displacement returns to zero: it ends when the displacement equals minus the height. Choose up as positive and keep the sign.

Worked example 3.6 · Thrown up from a cliff

A stone is thrown at $15\text{ m s}^{-1}$ at $30^\circ$ above the horizontal from the top of a cliff $30$ m above the sea. Find how far from the foot of the cliff it enters the water.

Vertically, with up positive, it ends $30$ m below its start:
$\begin{aligned} -30 &= 7.5t - 4.9t^2 \\ 4.9t^2 - 7.5t - 30 &= 0 \end{aligned}$
The positive root is $t = 3.355\ldots$ s.
Horizontally: $15\cos 30^\circ \times 3.355\ldots = 43.6$ m (3 s.f.).

Practice

Twelve questions in three tiers. Try each one properly before opening the solution. The solutions show the full method, not just answers. Take $g = 9.8\text{ m s}^{-2}$ , ignore air resistance, and give answers to 3 significant figures where they aren’t exact.

Warm-up

1 [3 marks]

A stone is thrown horizontally at $8\text{ m s}^{-1}$ from a point $45$ m above level ground. Find the time it takes to land and how far it travels horizontally.

Show solution

Vertically: $45 = 4.9t^2$ , so $t = 3.03$ s. Horizontally: $8 \times 3.030\ldots = 24.2$ m.

2 [2 marks]

A particle is projected at $30\text{ m s}^{-1}$ at $50^\circ$ above the horizontal. Find the horizontal and vertical components of its initial velocity.

Show solution

Horizontal: $30\cos 50^\circ = 19.3\text{ m s}^{-1}$ . Vertical: $30\sin 50^\circ = 23.0\text{ m s}^{-1}$ .

3 [4 marks]

A ball is projected from level ground at $20\text{ m s}^{-1}$ at $30^\circ$ above the horizontal. Find its time of flight and its range.

Show solution

Vertically: $0 = 10t - 4.9t^2$ , so $t = \dfrac{10}{4.9} = 2.04$ s.

Horizontally: $20\cos 30^\circ \times 2.040\ldots = 35.3$ m.

4 [3 marks]

A particle is projected at $18\text{ m s}^{-1}$ at $60^\circ$ above the horizontal. Find its greatest height above the point of projection.

Show solution

At the top $v_y = 0$ : $0 = (18\sin 60^\circ)^2 - 19.6H$ , so $H = \dfrac{243}{19.6} = 12.4$ m.

Standard

5 [5 marks]

A ball is projected at $20\text{ m s}^{-1}$ at $35^\circ$ above the horizontal. Find its speed and direction of motion $1.5$ s later.

Show solution

\begin{aligned} v_x &= 20\cos 35^\circ = 16.38\ldots \\ v_y &= 20\sin 35^\circ - 9.8 \times 1.5 \\ &= -3.228\ldots \end{aligned}

The speed is $\sqrt{v_x^2 + v_y^2} = 16.7\text{ m s}^{-1}$ . $v_y$ is negative, so the ball is falling, at $\tan^{-1}\tfrac{3.228\ldots}{16.38\ldots} = 11.1^\circ$ below the horizontal.

6 [4 marks]

A particle is projected from level ground at speed $U$ at an angle $\alpha$ above the horizontal. Show that its greatest height is $\dfrac{U^2\sin^2\alpha}{2g}$ .

Show solution

Vertically the initial velocity is $U\sin\alpha$ and the acceleration is $-g$ . At the greatest height the vertical velocity is zero, so by $v^2 = u^2 + 2as$ :

\begin{aligned} 0 &= U^2\sin^2\alpha - 2gH \\ H &= \frac{U^2\sin^2\alpha}{2g} \end{aligned}

7 [5 marks]

A ball is struck from level ground at $25\text{ m s}^{-1}$ and lands $40$ m away. Find the two possible angles of projection.

Show solution

The range is $\dfrac{U^2\sin 2\alpha}{g}$ , so

\sin 2\alpha = \frac{40 \times 9.8}{625} = 0.6272

$2\alpha = 38.84\ldots^\circ$ or $180^\circ - 38.84\ldots^\circ$ , so $\alpha = 19.4^\circ$ or $\alpha = 70.6^\circ$ : a low drive or a high lob.

8 [4 marks]

A marble rolls off a horizontal table $1.2$ m high and lands on the floor $1.5$ m from the edge of the table. Find the speed of the marble as it leaves the table.

Show solution

Vertically: $1.2 = 4.9t^2$ , so $t = 0.4948\ldots$ s. Horizontally the speed is constant: $u = \dfrac{1.5}{0.4948\ldots} = 3.03\text{ m s}^{-1}$ .

9 [4 marks]

A particle is projected from the origin at $10\text{ m s}^{-1}$ at $45^\circ$ above the horizontal. Find the equation of its path.

Show solution

$\tan 45^\circ = 1$ and $\cos^2 45^\circ = \tfrac{1}{2}$ , so

\begin{aligned} y &= x - \frac{9.8x^2}{2 \times 100 \times \frac{1}{2}} \\ &= x - 0.098x^2 \end{aligned}

Exam style

10 [9 marks]

A golf ball is hit from level ground at $30\text{ m s}^{-1}$ at an angle $\alpha$ above the horizontal, where $\tan\alpha = \tfrac{3}{4}$ .

Find the time of flight. [2]
Find the range. [2]
Find the greatest height. [2]
Find for how long the ball is more than $10$ m above the ground. [3]

Show solution

$\sin\alpha = \tfrac{3}{5}$ and $\cos\alpha = \tfrac{4}{5}$ , so the components are $18$ vertically and $24$ horizontally.

(a) $0 = 18t - 4.9t^2$ , so $t = \dfrac{18}{4.9} = 3.67$ s.

(b) $24 \times 3.673\ldots = 88.2$ m.

(c) $0 = 18^2 - 19.6H$ , so $H = 16.5$ m.

(d) Solve $18t - 4.9t^2 = 10$ , that is $4.9t^2 - 18t + 10 = 0$ :

\begin{aligned} t &= \frac{18 \pm \sqrt{128}}{9.8} \\ &= 0.682\ldots \text{ or } 2.991\ldots \end{aligned}

The ball is above $10$ m between these times, for $2.31$ s.

11 [8 marks]

A stone is thrown at $20\text{ m s}^{-1}$ at $30^\circ$ above the horizontal from the top of a cliff $25$ m above the sea.

Find the time taken for the stone to reach the sea. [4]
Find how far from the foot of the cliff it enters the water. [2]
Find its speed as it enters the water. [2]

Show solution

(a) Vertically, with up positive, $u = 10$ and the displacement is $-25$ :

\begin{aligned} -25 &= 10t - 4.9t^2 \\ 4.9t^2 - 10t - 25 &= 0 \end{aligned}

The positive root is $t = \dfrac{10 + \sqrt{590}}{9.8} = 3.50$ s.

(b) $20\cos 30^\circ \times 3.498\ldots = 60.6$ m.

(c) $v_x = 17.32\ldots$ and $v_y = 10 - 9.8 \times 3.498\ldots = -24.28\ldots$ , so the speed is $\sqrt{v_x^2 + v_y^2} = 29.8\text{ m s}^{-1}$ .

12 [7 marks]

A ball is kicked from ground level at $20\text{ m s}^{-1}$ at $30^\circ$ above the horizontal, towards a wall $3$ m high that is $15$ m away.

Show that the ball passes over the wall. [4]
Find how far beyond the wall the ball lands. [3]

Show solution

(a) The path is $y = x\tan 30^\circ - \dfrac{9.8x^2}{2 \times 400\cos^2 30^\circ}$ . At $x = 15$ :

y = 8.660\ldots - 3.675 = 4.99 \text{ m}

That is more than $3$ m, so the ball clears the wall.

(b) The range is $\dfrac{400\sin 60^\circ}{9.8} = 35.3$ m, so it lands $35.3 - 15 = 20.3$ m beyond the wall.