3.02MechanicsStretch

Projectile Motion

Q: What angle gives the maximum range?

45^ on level ground, since 2 is greatest when 2 = 90^.

A ball kicked across a pitch traces a curved path, but the mathematics is simpler than it looks. Ignoring air resistance, the only force is gravity, which acts vertically. So horizontally the ball moves at constant velocity, vertically it moves with constant acceleration $g$ downwards, and the two motions happen independently — linked only by the fact that they take the same time.

50 min Kinematics

The big picture

Projectile motion is the payoff for treating velocity as a vector. Resolve the launch velocity into horizontal and vertical components, and a two-dimensional problem becomes two one-dimensional ones you already know how to solve: constant velocity across, suvat up and down. Time is the variable common to both, so the standard strategy is to find $t$ from one direction and use it in the other. From that come the classic results — time of flight, range, greatest height — and the Cartesian equation of the trajectory, which shows the path is a parabola. Being able to derive these, rather than quote them, is what OCR tests, along with a clear statement of the modelling assumptions.

What you'll be able to do

Resolve a launch velocity into horizontal and vertical components
Model horizontal motion as constant velocity and vertical motion as constant acceleration
Find time of flight, range, greatest height, and velocity at any time
Derive the equation of the trajectory and state the modelling assumptions

Resolving the initial velocity

A projectile launched at speed $U$ at angle $\theta$ above the horizontal has horizontal component $U\cos\theta$ and vertical component $U\sin\theta$ .

Model it as a particle, with no air resistance, moving under gravity alone. Then the horizontal acceleration is $0$ and the vertical acceleration is $-g$ (up positive).

Horizontal: $x = (U\cos\theta)t$ . Vertical: $y = (U\sin\theta)t - \tfrac12gt^{2}$ and $v_{y} = U\sin\theta - gt$ .

x = (U\cos\theta)\,t

Horizontal: constant velocity.

y = (U\sin\theta)\,t - \tfrac12gt^{2}

Vertical: constant acceleration.

Tip — Write two separate columns, "horizontal" and "vertical", with $s, u, v, a, t$ in each. Time is the only quantity shared by both columns.

Time of flight, range and greatest height

The $\textbf{time of flight}$ over level ground comes from $y = 0$ : $t = \dfrac{2U\sin\theta}{g}$ .

The $\textbf{range}$ is the horizontal distance in that time: $R = \dfrac{U^{2}\sin 2\theta}{g}$ , which is greatest at $\theta = 45^{\circ}$ .

The $\textbf{greatest height}$ is where the vertical velocity is zero: $H = \dfrac{U^{2}\sin^{2}\theta}{2g}$ .

Example 1A ball is kicked from level ground at 25 m s⁻¹ at

40^{\circ}

to the horizontal. Find the time of flight, range and greatest height. Take

g = 9.8

m s⁻².

1Components:

u_{x} = 25\cos 40^{\circ} = 19.15

and

u_{y} = 25\sin 40^{\circ} = 16.07

2Lands when

16.07t - 4.9t^{2} = 0

t = \dfrac{16.07}{4.9} = 3.28

3Range

= 19.15 \times 3.28 = 62.8

4Greatest height:

0 = 16.07^{2} - 19.6H

, so

H = 13.2

AnswerTime

3.28

s, range

62.8

m, greatest height

13.2

Why this works

Because $\sin 2\theta = \sin(180^{\circ} - 2\theta)$ , complementary angles such as $30^{\circ}$ and $60^{\circ}$ give the same range. The steeper shot goes higher and takes longer to cover the same distance.

Launching from a height, and velocity during flight

If a projectile starts above the ground, the landing condition is $y = -h$ , not $y = 0$ . Solve the quadratic for $t$ and reject the negative root.

At any time the velocity has components $v_{x} = U\cos\theta$ and $v_{y} = U\sin\theta - gt$ . Its speed is $\sqrt{v_{x}^{2} + v_{y}^{2}}$ and its direction is $\tan^{-1}\dfrac{v_{y}}{v_{x}}$ to the horizontal.

Example 1A stone is thrown horizontally at 18 m s⁻¹ from the top of a 45 m cliff. Find how far from the foot of the cliff it lands, and its speed and direction on impact.

1Vertical:

u_{y} = 0

s = -45

-45 = -4.9t^{2}

, so

t = 3.030

2Horizontal:

x = 18 \times 3.030 = 54.5

3At impact

v_{y} = -9.8 \times 3.030 = -29.7

v_{x} = 18

4Speed

= \sqrt{18^{2} + 29.7^{2}} = 34.7

; angle

= \tan^{-1}\dfrac{29.7}{18} = 58.8^{\circ}

below the horizontal.

Answer

54.5

m from the cliff;

34.7

m s⁻¹ at

58.8^{\circ}

below the horizontal.

Tip — When asked for velocity, give both a magnitude and a direction. A speed alone answers only half the question.

The trajectory equation

Eliminating $t$ gives the path. From $x = (U\cos\theta)t$ , $t = \dfrac{x}{U\cos\theta}$ . Substituting into the vertical equation gives the trajectory below — a parabola.

Using $\sec^{2}\theta = 1 + \tan^{2}\theta$ , it can also be written in terms of $\tan\theta$ , which is how "find the angle needed to hit a target" questions become quadratics in $\tan\theta$ .

y = x\tan\theta - \dfrac{gx^{2}}{2U^{2}\cos^{2}\theta} = x\tan\theta - \dfrac{gx^{2}}{2U^{2}}(1 + \tan^{2}\theta)

Equation of the trajectory.

Example 1A ball kicked at 20 m s⁻¹ at

30^{\circ}

travels towards a wall 30 m away. How high is it when it reaches the wall?

1Horizontal:

30 = 17.32t

, so

t = 1.732

2Vertical:

y = 10 \times 1.732 - 4.9 \times 1.732^{2} = 17.32 - 14.70

Answer

2.62

m above the ground

Think like an examiner

State modelling assumptions: particle, no air resistance, constant $g$ .
Show resolved components explicitly.
Find time from one direction and use it in the other.
Give velocities with magnitude and direction.

Common misconceptions

Applying

g

to the horizontal motion.

Horizontal velocity is constant; gravity acts only vertically.

Using the launch speed in vertical suvat.

Use the vertical component

U\sin\theta

Setting

y = 0

for landing when launched from a height.

Use

y = -h

relative to the launch point.

Assuming velocity is zero at the top of the path.

Only the vertical component is zero; the horizontal component remains.

Projectiles

T = \tfrac{2U\sin\theta}{g}

Time of flight (level ground).

R = \tfrac{U^{2}\sin 2\theta}{g}

Range.

H = \tfrac{U^{2}\sin^{2}\theta}{2g}

Greatest height.

y = x\tan\theta - \tfrac{gx^{2}}{2U^{2}\cos^{2}\theta}

Trajectory.

Stretch yourself

A projectile is launched from level ground at 21 m s⁻¹ and lands 40 m away. Find the two possible angles of projection, and the greatest height reached in each case. Take $g = 9.8$ m s⁻².

Hint — Use $R = \dfrac{U^{2}\sin 2\theta}{g}$ to find $\sin 2\theta$ , then find both solutions for $2\theta$ between $0^{\circ}$ and $180^{\circ}$ .

Reveal solution

Solution

40 = \dfrac{441\sin 2\theta}{9.8}

, so

\sin 2\theta = \dfrac{392}{441} = 0.8889

. Then

2\theta = 62.73^{\circ}

117.27^{\circ}

, giving

\theta = 31.4^{\circ}

58.6^{\circ}

. Greatest heights,

H = \dfrac{U^{2}\sin^{2}\theta}{2g}

: for

31.4^{\circ}

H = \dfrac{441 \times 0.2715}{19.6} = 6.11

m; for

58.6^{\circ}

H = \dfrac{441 \times 0.7285}{19.6} = 16.4

m. The two angles are complementary, as expected, and the steeper launch climbs much higher to reach the same point.

\blacksquare

Questions students ask

Why can horizontal and vertical motion be treated separately?

Gravity acts only vertically, so it changes only the vertical component of velocity. With no air resistance nothing affects the horizontal component.

What angle gives the maximum range?

45^{\circ}

on level ground, since

\sin 2\theta

is greatest when

2\theta = 90^{\circ}

Should I memorise the range and height formulas?

Know how to derive them — OCR often asks you to show them — but in most problems it is safer to work from components.

How does air resistance change the answers?

It reduces both the range and the greatest height, and makes the path asymmetric. The simple model overestimates both.

Key takeaways

Resolve: $U\cos\theta$ horizontally, $U\sin\theta$ vertically.
Horizontal motion has constant velocity; vertical motion has acceleration $-g$ .
Time links the two directions.
On level ground: $T = \tfrac{2U\sin\theta}{g}$ , $R = \tfrac{U^{2}\sin 2\theta}{g}$ , $H = \tfrac{U^{2}\sin^{2}\theta}{2g}$ .
The trajectory is a parabola; complementary angles give equal ranges.