Projectile motion

AS · 9 min

A ball kicked across a field, a stone thrown off a cliff and an electron fired between charged plates all follow the same rule: a uniform velocity in one direction combined with a uniform acceleration in the perpendicular direction. The key idea is that the two perpendicular motions are completely independent, so you solve each with the tools you already have and link them through time. Projectile questions are a staple of Paper 2 and appear in Paper 1 as quick calculations.

Independence of perpendicular motions

Drop one ball and, at the same instant, fire another horizontally from the same height. They hit the ground together. The horizontal velocity of the second ball has no effect on how fast it falls.

This works because the only force on a projectile (with air resistance neglected) is its weight, which acts vertically. So:

Horizontally: no force, so no acceleration. The horizontal velocity is constant.
Vertically: acceleration $g = 9.81\ \text{m s}^{-2}$ downwards, exactly as in free fall.

Key result

For a projectile launched at speed $u$ at angle $\theta$ above the horizontal (no air resistance):

	Horizontal	Vertical (up positive)
initial velocity	$u_x = u\cos\theta$	$u_y = u\sin\theta$
acceleration	$0$	$-g$
velocity at time $t$	$v_x = u\cos\theta$ (constant)	$v_y = u\sin\theta - gt$
displacement at time $t$	$x = (u\cos\theta)\,t$	$y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2$

The time $t$ is the same in both columns: it is the link between them.

The path is a parabola. Eliminating $t$ between the two displacement equations gives

y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}

which is a quadratic in $x$ . You do not need to memorise this; it is here to show why the shape is a parabola.

y = 0.7002 x - 0.01828 x^2

The trajectory of a ball kicked at $20\ \text{m s}^{-1}$ at $35^\circ$ to the horizontal. It is symmetric about the highest point, which is reached halfway through the flight.

Method

Solving a projectile problem

Resolve the initial velocity into horizontal and vertical components.
Choose a positive vertical direction (usually up) and write $a_y = -9.81\ \text{m s}^{-2}$ .
Use the vertical motion (suvat) to find the time: the time to the top ( $v_y = 0$ ), the time of flight (return to a given height), or the time to fall a given distance.
Use the horizontal motion with that time: $x = u_x t$ .
If you need the velocity at some point, find $v_x$ and $v_y$ separately, then combine with Pythagoras for the speed and $\tan^{-1}(v_y/v_x)$ for the angle.

Watch out

Never put the launch speed $u$ itself into a suvat equation for either direction. Each direction uses its own component. A frequent error is writing $s = ut + \tfrac{1}{2}at^2$ vertically with $u = 20\ \text{m s}^{-1}$ instead of $20\sin 35^\circ$ .

Horizontal launch

When an object is launched horizontally, the vertical component of its initial velocity is zero. The vertical motion is identical to an object dropped from rest.

Ball rolling off a table

A ball rolls off the edge of a horizontal table $0.80\ \text{m}$ high at $3.0\ \text{m s}^{-1}$ . Neglecting air resistance, calculate (a) the time to reach the floor, (b) the horizontal distance from the table edge to where it lands, and (c) its velocity just before landing.

Solution

(a) Vertically (down positive), $u_y = 0$ , $s = 0.80\ \text{m}$ , $a = 9.81\ \text{m s}^{-2}$ :

0.80 = \tfrac{1}{2}(9.81)t^2 \quad\Rightarrow\quad t = \sqrt{\frac{1.60}{9.81}} = 0.404\ \text{s}

(b) Horizontally, at constant velocity:

x = 3.0 \times 0.404 = 1.21\ \text{m}

v = \sqrt{3.0^2 + 3.96^2} = 4.97\ \text{m s}^{-1}, \qquad \tan\phi = \frac{3.96}{3.0} \Rightarrow \phi = 53^\circ \text{ below the horizontal}

Launch at an angle

Ball kicked from the ground

A football is kicked from level ground at $20\ \text{m s}^{-1}$ at $35^\circ$ above the horizontal. Neglecting air resistance, calculate (a) the time of flight, (b) the range (horizontal distance travelled) and (c) the maximum height.

Solution

Components: $u_x = 20\cos 35^\circ = 16.38\ \text{m s}^{-1}$ , $u_y = 20\sin 35^\circ = 11.47\ \text{m s}^{-1}$ . Up positive.

(a) The ball lands at the same height, so the vertical displacement is zero:

0 = 11.47t - 4.905t^2 \quad\Rightarrow\quad t = \frac{11.47}{4.905} = 2.34\ \text{s}

(b)

R = u_x t = 16.38 \times 2.34 = 38.3\ \text{m}

0 = 11.47^2 - 2(9.81)h \quad\Rightarrow\quad h = \frac{131.6}{19.62} = 6.71\ \text{m}

The top is reached at $t = 11.47/9.81 = 1.17\ \text{s}$ , exactly half the time of flight, because the path is symmetric on level ground.

At the highest point the velocity is horizontal and equal to $u_x$ . It is not zero: only the vertical component is zero. The acceleration is still $9.81\ \text{m s}^{-2}$ downwards there.

Tip

On level ground, the range is greatest for a launch angle of $45^\circ$ (with no air resistance), and complementary angles such as $30^\circ$ and $60^\circ$ give equal ranges. The range formula $R = \dfrac{u^2\sin 2\theta}{g}$ shows this, but it is not on the syllabus and only applies when launch and landing heights are equal; derive results from components instead.

Thrown from a cliff

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

Solution

$u_x = 15\cos 30^\circ = 12.99\ \text{m s}^{-1}$ , $u_y = 15\sin 30^\circ = 7.50\ \text{m s}^{-1}$ . Up positive; the sea is $30\ \text{m}$ below the launch point, so $s_y = -30\ \text{m}$ .

-30 = 7.50t - 4.905t^2 \quad\Rightarrow\quad 4.905t^2 - 7.50t - 30 = 0

t = \frac{7.50 + \sqrt{7.50^2 + 4(4.905)(30)}}{2(4.905)} = \frac{7.50 + 25.39}{9.81} = 3.35\ \text{s}

Horizontal distance $= 12.99 \times 3.35 = 43.6\ \text{m}$ .

Vertical velocity at impact: $v_y = 7.50 - 9.81(3.353) = -25.4\ \text{m s}^{-1}$ (downwards). Speed:

v = \sqrt{12.99^2 + 25.39^2} = 28.5\ \text{m s}^{-1}

Check with energy: $v^2 = u^2 + 2gh = 225 + 2(9.81)(30) = 813.6$ , so $v = 28.5\ \text{m s}^{-1}$ . The speed at impact does not depend on the launch angle.

Clearing a wall

A ball is thrown at $18\ \text{m s}^{-1}$ at $40^\circ$ above the horizontal from ground level towards a wall $3.0\ \text{m}$ high, $25\ \text{m}$ away. Does it clear the wall, and by how much?

Solution

$u_x = 18\cos 40^\circ = 13.79\ \text{m s}^{-1}$ , $u_y = 18\sin 40^\circ = 11.57\ \text{m s}^{-1}$ .

Time to reach the wall (horizontal motion): $t = 25/13.79 = 1.813\ \text{s}$ .

Height at that time (vertical motion, up positive):

y = 11.57(1.813) - 4.905(1.813)^2 = 20.98 - 16.12 = 4.86\ \text{m}

The ball is $4.9\ \text{m}$ high at the wall, so it clears it by about $1.9\ \text{m}$ . (Is it still rising? $v_y = 11.57 - 9.81(1.813) = -6.2\ \text{m s}^{-1}$ , so it is already on the way down.)

Other uniform-acceleration situations

The syllabus phrase is "uniform velocity in one direction and a uniform acceleration in a perpendicular direction". Gravity is the usual cause, but the same analysis applies whenever a constant force acts perpendicular to the initial velocity. A charged particle entering a uniform electric field between parallel plates (an A Level topic) moves in exactly this way: constant velocity along the plates, constant acceleration across them, parabolic path.

The effect of air resistance

With air resistance, the projectile experiences a drag force opposite to its velocity, which grows with speed. Compared with the ideal parabola:

the maximum height is lower;
the range is shorter;
the path is not symmetric: the descent is steeper than the ascent, because the horizontal velocity keeps decreasing;
the horizontal component of velocity is no longer constant; it decreases throughout the flight.

You are expected to describe these effects qualitatively (see Drag and terminal velocity), not calculate them.

Exam tip

Show the resolved components explicitly: " $u_y = 20\sin 35^\circ = 11.5\ \text{m s}^{-1}$ ". These are often method marks.
"Describe and explain the motion" of a projectile: state that the horizontal velocity is constant because there is no horizontal force (air resistance neglected), and the vertical motion has constant downward acceleration $g$ because the weight acts vertically.
At the highest point the speed is $u\cos\theta$ , not zero; Paper 1 tests this repeatedly, often through kinetic energy at the top: $E_K = \tfrac{1}{2}m(u\cos\theta)^2$ .
When asked for a velocity, give the magnitude and the angle (and say "below the horizontal" or "above").

Summary

Horizontal and vertical motions are independent; time links them.
Horizontally: constant velocity $u\cos\theta$ (no air resistance).
Vertically: uniform acceleration $g$ downwards, starting from $u\sin\theta$ .
Use the vertical motion to find time, then the horizontal motion to find distance.
Horizontal launch: vertical motion is the same as dropping from rest.
At the top, $v_y = 0$ but $v_x$ is unchanged; acceleration is still $g$ downwards.
Air resistance lowers the maximum height, shortens the range and makes the path asymmetric.

Practice

Question

A ball is thrown horizontally at $12\ \text{m s}^{-1}$ from a height of $1.8\ \text{m}$ . How far does it travel horizontally before hitting the ground?
An arrow is fired horizontally at $60\ \text{m s}^{-1}$ at a target $30\ \text{m}$ away, aimed at the bullseye. How far below the bullseye does it hit?
A projectile is launched at $25\ \text{m s}^{-1}$ at $60^\circ$ above the horizontal over level ground. Calculate its time of flight, range and maximum height.
For the projectile in question 3, calculate its speed at the highest point and at $t = 1.0\ \text{s}$ .
Explain why a bullet fired horizontally and a bullet dropped from the same height at the same instant hit level ground at the same time (neglect air resistance).
A stone is thrown horizontally from a cliff and hits the sea $2.5\ \text{s}$ later, $35\ \text{m}$ from the foot of the cliff. Find the height of the cliff and the speed of projection.
A ball leaves the ground at $15\ \text{m s}^{-1}$ at $50^\circ$ to the horizontal. Find the two times at which it is $5.0\ \text{m}$ above the ground.
Sketch, on the same axes, the paths of a projectile launched at the same speed and angle (a) without and (b) with air resistance. State three differences.
A plane flying horizontally at $80\ \text{m s}^{-1}$ at a height of $200\ \text{m}$ releases a food package. How far before the drop zone (measured horizontally) must the package be released? What is its velocity on landing?
A ball is thrown from the edge of a roof $12\ \text{m}$ high with speed $10\ \text{m s}^{-1}$ at $20^\circ$ below the horizontal. Find the time to reach the ground and the horizontal distance travelled.

Answers

$t = \sqrt{2(1.8)/9.81} = 0.606\ \text{s}$ ; $x = 12 \times 0.606 = 7.3\ \text{m}$ .
$t = 30/60 = 0.50\ \text{s}$ ; drop $= \tfrac{1}{2}(9.81)(0.50)^2 = 1.23\ \text{m}$ .
$u_x = 12.5\ \text{m s}^{-1}$ , $u_y = 21.65\ \text{m s}^{-1}$ . $t = 2u_y/g = 4.41\ \text{s}$ ; range $= 12.5 \times 4.41 = 55.2\ \text{m}$ ; $h = 21.65^2/19.62 = 23.9\ \text{m}$ .
At the top, $v = u_x = 12.5\ \text{m s}^{-1}$ . At $1.0\ \text{s}$ : $v_y = 21.65 - 9.81 = 11.84\ \text{m s}^{-1}$ , $v = \sqrt{12.5^2 + 11.84^2} = 17.2\ \text{m s}^{-1}$ .
Vertically, both start with zero vertical velocity and have the same downward acceleration $g$ (the only force, weight, is vertical). The horizontal velocity does not affect the vertical motion, so both take the same time to fall the same height.
Height $= \tfrac{1}{2}(9.81)(2.5)^2 = 30.7\ \text{m}$ ; speed $= 35/2.5 = 14\ \text{m s}^{-1}$ .
$u_y = 15\sin 50^\circ = 11.49\ \text{m s}^{-1}$ . $5.0 = 11.49t - 4.905t^2$ , so $4.905t^2 - 11.49t + 5.0 = 0$ , giving $t = \dfrac{11.49 \pm \sqrt{132.0 - 98.1}}{9.81} = \dfrac{11.49 \pm 5.82}{9.81}$ : $t = 0.58\ \text{s}$ (rising) and $t = 1.76\ \text{s}$ (falling).
With air resistance: lower maximum height; shorter range; asymmetric path with a steeper descent than ascent (the highest point is more than halfway along the range); horizontal velocity decreasing rather than constant. Any three.
$t = \sqrt{2(200)/9.81} = 6.39\ \text{s}$ ; release $80 \times 6.39 = 511\ \text{m}$ before the zone. $v_y = 9.81 \times 6.39 = 62.6\ \text{m s}^{-1}$ ; $v = \sqrt{80^2 + 62.6^2} = 102\ \text{m s}^{-1}$ at $\tan^{-1}(62.6/80) = 38^\circ$ below the horizontal.
Down positive: $u_y = 10\sin 20^\circ = 3.42\ \text{m s}^{-1}$ , $u_x = 9.40\ \text{m s}^{-1}$ . $12 = 3.42t + 4.905t^2$ , so $t = \dfrac{-3.42 + \sqrt{11.70 + 235.4}}{9.81} = 1.25\ \text{s}$ . Distance $= 9.40 \times 1.25 = 11.8\ \text{m}$ .

Independence of perpendicular motions

Horizontal launch

Launch at an angle

Other uniform-acceleration situations

The effect of air resistance

Summary

Practice

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