Two-Dimensional & Projectile Motion

The trick that makes it easy

Roll a ball off the edge of a table and watch it. It keeps moving forward while it falls, and the path it traces is a smooth curve. Now hold a second ball at the table edge and drop it at the exact moment the first one leaves. They hit the floor together. Not roughly together: together, every time, no matter how fast you rolled the first one.

That is the whole subject in one experiment. Moving sideways does nothing to how fast you fall. Falling does nothing to how fast you move sideways. So a problem that looks two-dimensional and frightening is really two one-dimensional problems you already solved in the last lesson, running side by side and sharing a single clock.

Two equations and one clock

Set up axes with $x$ horizontal and $y$ vertical, positive upward, with the origin at the launch point. Call the launch speed $v_0$ and the launch angle $\theta$, measured up from the horizontal. The velocity at launch splits into

$v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta$

Horizontally, nothing pushes the projectile once it is in the air, so $a_x = 0$ and the horizontal motion is the uniform motion you met first: position grows in equal steps in equal times. Vertically, gravity pulls down with constant $a_y = -g$, where $g = 9.8\ \mathrm{m/s^2}$ near Earth's surface (many contest problems say to use $10$, and will tell you). Feed each of those into the constant-acceleration equations from the previous lesson and you get the pair that runs everything:

$x(t) = v_0\cos\theta\; t, \qquad y(t) = v_0\sin\theta\; t - \tfrac{1}{2}gt^2$

$v_x(t) = v_0\cos\theta, \qquad v_y(t) = v_0\sin\theta - gt$

Read those in words. The horizontal position climbs steadily and the horizontal velocity never changes, not at the top, not on the way down, not ever. The vertical velocity starts positive, shrinks at $9.8\ \mathrm{m/s}$ every second, passes through zero at the top of the flight, then goes negative as the projectile comes back down. The two lines never refer to each other. The only thing they share is $t$.

That sharing is the engine of every projectile problem. One direction tells you the time; you hand that time to the other direction.

Worked example 1: off a cliff

A ball is thrown horizontally at $15\ \mathrm{m/s}$ from the edge of a $20\ \mathrm{m}$ cliff. How long is it in the air, and how far from the base does it land?

On blank paper, write the two equations with the numbers in. Horizontally the launch is level, so $\theta = 0$, giving $v_{0x} = 15\ \mathrm{m/s}$ and $v_{0y} = 0$. Put the origin at the launch point, so the ground is at $y = -20\ \mathrm{m}$.

The vertical line has one unknown, so start there:

$-20 = 0\cdot t - \tfrac{1}{2}(9.8)t^2 \;\Rightarrow\; t^2 = \frac{40}{9.8} = 4.082 \;\Rightarrow\; t = 2.02\ \mathrm{s}$

Now hand that time across:

$x = 15 \times 2.02 = 30.3\ \mathrm{m}$

The horizontal speed never entered the time calculation, and gravity never entered the distance calculation. If you doubled the throwing speed, the ball would still take $2.02\ \mathrm{s}$ and would land twice as far out.

Level ground: three formulas you can derive in ten seconds

When the projectile lands at the same height it started from, the algebra is tidy enough to be worth carrying in your head. Landing means $y = 0$ again:

$0 = v_0\sin\theta\, t - \tfrac{1}{2}gt^2 = t\left(v_0\sin\theta - \tfrac{1}{2}gt\right)$

One root is $t = 0$, the launch. The other is the landing, so the total time of flight is

$T = \frac{2v_0\sin\theta}{g}$

Put half of that into the vertical velocity and you find $v_y = 0$ there: the top of the flight sits exactly halfway through in time, and therefore exactly halfway along in horizontal distance, because $x$ grows steadily. Feeding $T$ into $x(t)$ and $T/2$ into $y(t)$ gives the other two:

$R = \frac{v_0^2\sin 2\theta}{g}, \qquad H = \frac{v_0^2\sin^2\theta}{2g}$

using $2\sin\theta\cos\theta = \sin 2\theta$. Two consequences get tested constantly. First, $\sin 2\theta$ is largest when $2\theta = 90^\circ$, so the range peaks at $\theta = 45^\circ$ with $R_{\max} = v_0^2/g$. Second, $\sin 2\theta$ is symmetric about that peak: $\sin(2 \times 25^\circ)$ equals $\sin(2 \times 65^\circ)$. Any two angles that add to $90^\circ$ give the same range at the same speed. The steep one takes longer and climbs higher; it just lands in the same place.

One more relation is worth deriving because it turns up disguised. Divide $H$ by $R$:

$\frac{H}{R} = \frac{v_0^2\sin^2\theta/2g}{2v_0^2\sin\theta\cos\theta/g} = \frac{\tan\theta}{4} \;\Rightarrow\; \tan\theta = \frac{4H}{R}$

So if a problem tells you the range and the greatest height, the angle follows immediately and the speed follows from either formula. A problem that says "it passes $4.0\ \mathrm{m}$ above the ground at the horizontal midpoint of its flight" is telling you $H = 4.0\ \mathrm{m}$ in disguise, because the midpoint is the top.

Worked example 2: launched from a height

A projectile leaves a $15\ \mathrm{m}$ high platform at $35\ \mathrm{m/s}$ at $25^\circ$ above the horizontal. How far from the base of the platform does it land?

The landing is $15\ \mathrm{m}$ below the launch, so the range formula is out. Components first:

$v_{0x} = 35\cos 25^\circ = 31.72\ \mathrm{m/s}, \qquad v_{0y} = 35\sin 25^\circ = 14.79\ \mathrm{m/s}$

Origin at the launch point, ground at $y = -15\ \mathrm{m}$:

$-15 = 14.79\,t - 4.9\,t^2 \;\Rightarrow\; 4.9t^2 - 14.79t - 15 = 0$

That is a quadratic, so use the formula. The discriminant is

$14.79^2 + 4(4.9)(15) = 218.7 + 294 = 512.7, \qquad \sqrt{512.7} = 22.64$

$t = \frac{14.79 + 22.64}{9.8} = \frac{37.43}{9.8} = 3.820\ \mathrm{s}$

The other root is negative (it is the time the projectile would have left the ground had it been launched from there) so discard it. Then

$x = 31.72 \times 3.820 = 121.2\ \mathrm{m}$

Take the positive root every time the projectile lands below its launch point. Take the smaller positive root when a problem asks when the projectile first reaches a height it passes twice.

When the ground is tilted

A slope is where most people lose marks, and there are two clean ways through.

Method one: use the slope as a condition. A slope through the launch point at angle $\phi$ above the horizontal is the line $y = x\tan\phi$. Landing means the projectile's own $x$ and $y$ satisfy that line at the same instant. So set

$v_0\sin\theta\, t - \tfrac{1}{2}gt^2 = \left(v_0\cos\theta\, t\right)\tan\phi$

Cancel one factor of $t$ (that discarded root is the launch) and solve:

$t = \frac{2v_0\left(\sin\theta - \cos\theta\tan\phi\right)}{g}$

Then $x = v_0\cos\theta\, t$, and the distance measured along the slope, which is what these problems ask for, is $L = x/\cos\phi$. For a slope that descends, put $\phi$ in as a negative angle and everything still works.

Method two: tilt your axes. Point one axis along the slope and the other perpendicular to it. Gravity then has components $-g\sin\phi$ along the slope and $-g\cos\phi$ perpendicular to it. Landing is now "perpendicular displacement returns to zero", which is the level-ground problem again with $g\cos\phi$ in place of $g$ and the launch angle measured from the slope. Tilted axes are the faster route when the launch is described relative to the slope: "fired perpendicular to the surface", for instance.

Worked example 3: range up an incline

A projectile is fired from the bottom of a $20^\circ$ incline at $25\ \mathrm{m/s}$, at $50^\circ$ above the horizontal. How far up the slope does it land, measured along the slope?

Here $\phi = 20^\circ$ and $\theta = 50^\circ$, with $\tan 20^\circ = 0.3640$, $\sin 50^\circ = 0.7660$ and $\cos 50^\circ = 0.6428$. Time of flight:

$t = \frac{2(25)\left(0.7660 - 0.6428 \times 0.3640\right)}{9.8} = \frac{50\left(0.7660 - 0.2340\right)}{9.8} = \frac{50 \times 0.5320}{9.8} = 2.714\ \mathrm{s}$

Horizontal distance:

$x = 25 \times 0.6428 \times 2.714 = 43.62\ \mathrm{m}$

And along the slope:

$L = \frac{43.62}{\cos 20^\circ} = \frac{43.62}{0.9397} = 46.4\ \mathrm{m}$

If you want the best launch angle for a given slope, maximise that expression. It works out to $\theta = 45^\circ + \phi/2$ (split the difference between straight up the slope and the horizontal) and the maximum distance along the slope is

$L_{\max} = \frac{v_0^2}{g\left(1 + \sin\phi\right)}$

For $v_0 = 30\ \mathrm{m/s}$ up a $15^\circ$ slope that is $900 / \left[9.8(1.2588)\right] = 73.0\ \mathrm{m}$, achieved by aiming $52.5^\circ$ above the horizontal. Downhill, $\phi$ is negative, the denominator shrinks and you aim flatter.

Two optimisation results worth knowing

Greatest distance when you launch from a height. Throw from height $h$ with speed $v_0$ and the best angle is no longer $45^\circ$: the extra fall time rewards a flatter, faster throw. Maximising the landing $x$ over $\theta$ gives

$R_{\max} = \frac{v_0}{g}\sqrt{v_0^2 + 2gh}$

A shot put released $20\ \mathrm{m}$ up at $25\ \mathrm{m/s}$ reaches $\frac{25}{9.8}\sqrt{625 + 392} = 2.551 \times 31.89 = 81.4\ \mathrm{m}$. Check the limit: with $h = 0$ this collapses to $v_0^2/g$, the level-ground answer.

Smallest speed that reaches a given target. For a target at horizontal distance $x$ and height $y$, the minimum launch speed is

$v_{\min}^2 = g\left(y + \sqrt{x^2 + y^2}\right)$

For a target $30\ \mathrm{m}$ away and $10\ \mathrm{m}$ up, $\sqrt{900 + 100} = 31.62$, so $v_{\min}^2 = 9.8(41.62) = 407.9$ and $v_{\min} = 20.2\ \mathrm{m/s}$. The aiming angle that achieves it bisects the angle between straight up and the line to the target, which is the same "split the difference" rule as the slope result: a target on a slope is a slope problem.

The traps

Using $R = v_0^2\sin 2\theta/g$ when the heights differ. The single most common lost mark. Ask "does it land where it started, vertically?" before you write that formula.

Mixing up which component does what. $\cos$ goes with horizontal only because $\theta$ is measured from the horizontal. If a problem measures the angle from the vertical, or from a slope, the sine and cosine swap. Draw the triangle.

Forgetting that distance along a slope is not horizontal distance. The answer they want is usually $L = x/\cos\phi$, which is larger than $x$.

Taking the wrong quadratic root. The negative root is the projectile's imaginary past. Keep the positive one, unless the projectile crosses the target height twice and the problem wants the first crossing.

Forgetting to square. Range goes as $v_0^2$, not $v_0$. Doubling the launch speed quadruples the range.

How to recognise it

You are looking at a projectile problem when something is released or launched and then nothing but gravity acts on it: no thrust, no friction, no rope, no contact. The giveaways in the wording: "thrown", "kicked", "fired", "launched", "neglect air resistance", "leaves the edge".

Once you know it is one, the sentence that tells you which method:

• Launch and landing at the same height, asking for range, height or flight time: use $R$, $H$, $T$ directly.
• A cliff, platform, well or raised target: components and a quadratic in $t$.
• Thrown horizontally: set $\theta = 0$, so the vertical drop alone gives the time.
• Any mention of an incline, hillside, ramp or roof: the slope condition $y = x\tan\phi$, or tilted axes if the launch is described relative to the surface.
• "Greatest", "maximum", "minimum speed", "just barely reaches": an optimisation, and the answer usually bisects an angle.
• Range and maximum height both given: $\tan\theta = 4H/R$ first, then the speed.

The universal fallback, when nothing above fits, is always available: write $x(t)$ and $y(t)$, write down the condition the problem states, and solve. It is slower than a remembered formula and it has never once given a wrong answer.

Everything here assumed you were standing still watching the flight. Next you will ask what the same parabola looks like to someone on a moving train, which turns out to change the shape of the path but not the physics, in Relative Motion.