Lesson 1 of 5Dynamics

Unit 1 · Lesson 1 of 5

Motion in Two Dimensions and Projectiles

Splitting motion into perpendicular components, adding velocities between moving frames like rivers and wind, and solving any projectile by treating its horizontal and vertical motion as two separate problems joined only by time.

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Key things to remember

1Components: $v_{x} = v cos θ$ and $v_{y} = v sin θ$ ; recombine with $v = v_{x}^{2} + v_{y}^{2}$ and $θ = tan^{- 1} (v_{y} / v_{x})$ .
2Relative velocity adds as vectors: $v_{BG} = v_{B W} + v_{W G}$ .
3For a projectile with no air resistance, $a_{x} = 0$ and $a_{y} = 9.81 m/s^{2}$ downward; time is the only link between the two directions.
4At maximum height $v_{y} = 0$ , but $v_{x}$ is unchanged and the acceleration is still $9.81 m/s^{2}$ down.
5Level ground only: $R = \frac{v ^{2} sin 2 θ}{g}$ , maximum at 45°, equal for complementary angles.

In Grade 11 almost every motion ran along one line, so a plus or minus sign was enough to show direction. Real motion is rarely that tidy. A basketball arcs toward the hoop, a canoe drifts sideways while it crosses a river, a plane crabs into a crosswind. The good news is that you do not need new physics for any of these. You need one idea: motion in two perpendicular directions can be analyzed separately, then put back together at the end.

Components: one 2-D problem becomes two 1-D problems

Any vector can be replaced by two perpendicular components. A velocity of magnitude $v$ at an angle $θ$ above the horizontal has a horizontal component $v_{x} = v cos θ$ and a vertical component $v_{y} = v sin θ$ . Going the other way, the magnitude is $v = v_{x}^{2} + v_{y}^{2}$ and the angle is $θ = tan^{- 1} (v_{y} / v_{x})$ . Directions on a map are written as bearings like [N 35° E]: start facing north, then turn 35° toward east.

Independence of horizontal and vertical motion

Perpendicular components do not affect each other. For a projectile with no air resistance, the only force is gravity, which is vertical. So the horizontal velocity never changes (

a_{x} = 0

), while the vertical velocity changes at

a_{y} = 9.81 m/s^{2}

downward, exactly as it would for an object dropped straight down. The only quantity the two directions share is time.

Fig 1The horizontal component stays the same length at every point. The vertical component shrinks, reaches zero at the top, then grows downward.

Horizontal: Δ x = v_{x} Δ t Vertical: Δ y = v_{i y} Δ t + \frac{1}{2} a_{y} (Δ t)^{2}, v_{f y} = v_{i y} + a_{y} Δ t, v_{f y}^{2} = v_{i y}^{2} + 2 a_{y} Δ y

With up as positive, the vertical acceleration is negative 9.81 m/s squared. The time interval is the same in both rows.

Relative motion in two dimensions

Every velocity is measured relative to something. A boat's speedometer reads its velocity relative to the water; if the water itself moves relative to the ground, the two add as vectors. Write it with subscripts so the frames chain together: $v_{BG} = v_{B W} + v_{W G}$ (boat relative to ground equals boat relative to water plus water relative to ground). Planes work the same way, with air speed and wind in place of the boat and current.

Fig 2A boat aimed straight across a flowing river does not travel straight across.

AAim straight across

Heading is perpendicular to the banks
Crossing time uses the full boat speed: width divided by boat speed
Shortest possible crossing time
The current carries the boat downstream while it crosses

BArrive straight across

Heading is angled upstream so the current's push is cancelled
Only part of the boat's velocity points across the river
Takes longer than aiming straight across
Only possible if the boat is faster than the current

Three kinds of projectile

Launch	Starting components	What to find first
Horizontal, from a height $h$	$v_{i x} = v$ , $v_{i y} = 0$	Time from the vertical drop: $Δ t = 2 h / g$
At an angle, landing at launch height	$v_{i x} = v cos θ$ , $v_{i y} = v sin θ$	Time from $v_{f y} = - v_{i y}$ , so $Δ t = 2 v_{i y} / g$
At an angle, landing below launch height	$v_{i x} = v cos θ$ , $v_{i y} = v sin θ$	Time from $Δ y = v_{i y} Δ t + \frac{1}{2} a_{y} (Δ t)^{2}$ , a quadratic

At the top, the ball is still moving

At maximum height the vertical velocity is zero, but the horizontal velocity is exactly what it was at launch. The acceleration at the top is still

9.81 m/s^{2}

downward. Gravity never switches off partway through the flight.

Loading interactive…

Change the speed, angle and launch height, then scrub through time to watch the horizontal component stay fixed while the vertical one changes.

The level-ground range shortcut

Only when a projectile lands at the height it was launched from, the range is

R = \frac{v ^{2} sin 2 θ}{g}

. It is largest at 45°, and complementary angles such as 30° and 60° give the same range. From a cliff or onto a platform this formula does not apply; go back to components.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A boat can travel at

4.0 m/s

relative to the water. It heads straight across a river [N] that is

120 m

wide, while the current flows at

3.0 m/s

[E]. Find the boat's velocity relative to the ground, the crossing time, and how far downstream it lands.

Example 2

A marble rolls off a lab bench

1.20 m

high at

2.5 m/s

. How far from the base of the bench does it land, and what is its velocity just before it hits the floor?

Example 3

A soccer ball is kicked from level ground at

22 m/s

at 35° above the horizontal. Find its time of flight, maximum height and range.

Example 4

A stone is thrown from the top of a

20.0 m

cliff at

15.0 m/s

, 30.0° above the horizontal. How far from the base of the cliff does it land?

Where marks get lost

Using $R = v^{2} sin 2 θ / g$ for a launch from a cliff or onto a raised target. It only works when the landing height equals the launch height.
Saying the velocity is zero at the top of the path. Only the vertical component is zero; the horizontal component is unchanged.
Letting the horizontal velocity slow down during the flight. With no air resistance there is no horizontal force, so $v_{x}$ is constant.
Mixing a sign convention: taking up as positive for $v_{i y}$ but then using $a_{y} = + 9.81 m/s^{2}$ .
Using the full boat speed as the crossing speed when the boat is angled upstream. Only the component perpendicular to the banks carries it across.

Flashcards 7

Components of a vector at angle theta above horizontal

Projectile accelerations (no air resistance)

What links the horizontal and vertical motion of a projectile?

Relative velocity equation

+ 3 more

Practice

At the same instant, one ball is dropped from a height and an identical ball is fired horizontally from the same height. Ignoring air resistance, which lands first?

A ball rolls off a

0.90 m

high bench at

3.2 m/s

. How far from the base of the bench does it land, in metres? Give three significant figures.

A ball is kicked from level ground at

18 m/s

at 40° above the horizontal. What is its range, in metres? Give three significant figures.

A projectile is launched at

25 m/s

at 60° above the horizontal. What maximum height does it reach above the launch point, in metres? Give three significant figures.

A plane flies with an air speed of

250 km/h

heading [N] while the wind blows at

60 km/h

[E]. What is the plane's speed relative to the ground, in km/h? Give three significant figures.

km/h

A boat that moves at

5.0 m/s

relative to the water wants to arrive directly across a river flowing at

2.0 m/s

. At what angle upstream from straight across must it head, in degrees? Give three significant figures.

Ignoring air resistance, what is the acceleration of a thrown ball at the very top of its path?

On level ground, which launch angle gives the same range as 25° at the same speed?

For a projectile at its maximum height, what is the value of its vertical velocity (in m/s)?

Match each projectile quantity to how it behaves during the flight (no air resistance).

Horizontal velocity

Vertical velocity

Acceleration

Speed

Order the steps for solving a projectile launched at an angle from a cliff.

1Combine final components if the impact velocity is needed
2Use that time with the constant horizontal velocity to find the horizontal distance
3Solve the vertical motion for the time of flight
4Choose a positive direction and resolve the launch velocity into components
5List the vertical knowns: initial vertical velocity, displacement and acceleration

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