Vectors and Projectile Motion

In short: A projectile's horizontal and vertical motions are separate, and gravity only affects the vertical one. Launched at speed $u$ and angle $θ$ , $x = u t cos θ$ and $y = u t sin θ - \frac{1}{2} g t^{2}$ , and on flat ground the range is greatest at $4 5^{\circ}$ .

Starting point: what you already know

Vectors in $i$ , $j$ form, and $r cos θ i + r sin θ j$ for a vector of length $r$ at angle $θ$ .
From Mathematics Advanced: velocity is the derivative of displacement, $v = \frac{d x}{d t}$ , and acceleration is the derivative of velocity. Going backwards uses integration.
Gravity gives every falling object an acceleration of about $g = 9.8$ m/s² downward (often rounded to 10 in exam questions).
Solving quadratics and finding the maximum of a parabola.

Setting up the vectors

Put the origin at the launch point, $i$ horizontal and $j$ up. The only force (ignoring air resistance) is gravity, so

$a (t) = - g j$

Launch with speed $u$ at angle $θ$ to the horizontal:

$v (0) = u cos θ i + u sin θ j$

%%FIG:projectile%%

Integrate once for velocity and again for position (the constants come from the starting values):

Notice that the horizontal velocity $u cos θ$ never changes.

The three big questions

Use the free kick: $u = 20$ , $θ = 5 0^{\circ}$ , $g = 10$ . Then $u cos θ \approx 12.86$ and $u sin θ \approx 15.32$ .

1. Greatest height. At the top, the ball stops going up: vertical velocity is 0. $15.32 - 10 t = 0 \Rightarrow t \approx 1.53 s$ $y = 15.32 (1.53) - 5 (1.53)^{2} \approx 11.7 m$

2. Time of flight. It lands when $y = 0$ : $t (15.32 - 5 t) = 0 \Rightarrow t \approx 3.06 s$ This is twice the time to the top, as symmetry suggests.

3. Range. Horizontal distance at landing: $x = 12.86 \times 3.06 \approx 39.4 m$

Projectile motion

Set up

Origin at launch
a = -g j
v(0) = u cos θ i + u sin θ j

Equations

x = ut cos θ
y = ut sin θ - ½gt²

Questions

Top: vertical velocity 0
Flight time: y = 0
Range: x at landing

Path

Remove t
A parabola

Year 12General formulas and the best angle

Doing the same steps with letters: $time of flight T = \frac{2 u sin θ}{g}, greatest height H = \frac{u ^{2} sin ^{2} θ}{2 g}, range R = \frac{u ^{2} sin 2 θ}{g}$ Since $sin 2 θ$ is largest when $2 θ = 9 0^{\circ}$ , the longest range on flat ground comes from $θ = 4 5^{\circ}$ . Angles that add to $9 0^{\circ}$ , such as $3 0^{\circ}$ and $6 0^{\circ}$ , give the same range.

Hitting a target on level ground. Set the range equal to the distance: $\frac{u ^{2} sin 2 θ}{g} = d$ . With $u = 20$ , $g = 10$ and $d = 203$ : $sin 2 θ = \frac{3}{2}$ , so $2 θ = 6 0^{\circ}$ or $12 0^{\circ}$ , and $θ = 3 0^{\circ}$ or $6 0^{\circ}$ .

Passing through a point. Use the Cartesian path with $\frac{1}{cos ^{2} θ} = 1 + tan^{2} θ$ , which gives a quadratic in $T = tan θ$ . To pass through $(10, 5)$ with $u = 20$ and $g = 10$ : $5 = 10 T - \frac{10 \times 100}{800} (1 + T^{2})$ , so $1.25 T^{2} - 10 T + 6.25 = 0$ , that is $T^{2} - 8 T + 5 = 0$ and $T = 4 \pm 11$ . Two angles work.

Speed at the top. At the top only $u cos θ$ is left. If that is half the launch speed, $cos θ = \frac{1}{2}$ and $θ = 6 0^{\circ}$ .

Collisions. Two projectiles launched at the same moment from the same height collide only if they are always at the same height, so their vertical launch speeds must be equal: $u_{1} sin α = u_{2} sin β$ .

Up a slope. Given a formula such as the greatest range up a slope $\frac{u ^{2}}{g ( 1 + sin α )}$ , just substitute. Using the formula correctly is the skill being tested.

Quick summary

Horizontal and vertical motion are separate. Gravity only affects the vertical part.
$x = u t cos θ$ and $y = u t sin θ - \frac{1}{2} g t^{2}$ .
Greatest height: vertical velocity is 0. Flight time: $y = 0$ . Range: $x$ at landing.
On flat ground the range is $\frac{u ^{2} sin 2 θ}{g}$ , greatest at $4 5^{\circ}$ .

Quiz

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Try 3 questions now

Ignoring air resistance, the acceleration of a projectile is:
1. A zero
2. B $g$ downward
3. C $g$ forward
4. D it changes with time
Show answer
B: $g$ downward
Only gravity acts.
Which part of the velocity stays the same during flight?
1. A the horizontal part
2. B the vertical part
3. C both parts
4. D neither part
Show answer
A: the horizontal part
Gravity has no horizontal effect.
A ball is launched at 10 m/s at $3 0^{\circ}$ . Its initial vertical velocity is:
1. A 10 m/s
2. B 5 m/s
3. C $53$ m/s
4. D 0 m/s
Show answer
B: 5 m/s
$10 sin 3 0^{\circ} = 5$ .