Lesson 4 of 5Kinematics

Unit 1 · Lesson 4 of 5

Free Fall and Motion in Two Dimensions

Gravity as a constant acceleration of $9.81 m/s^{2}$ downward, then adding displacements and velocities that point in different directions.

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Key things to remember

1Free fall: $a = g = 9.81 m/s^{2}$ downward, the same for every mass, with air resistance ignored.
2Taking up as positive makes $a = - 9.81 m/s^{2}$ ; taking down as positive makes it $+ 9.81 m/s^{2}$ . State which you chose.
3At the top of a throw the velocity is zero but the acceleration is still $9.81 m/s^{2}$ downward.
4An object returning to its launch height has $Δ d = 0$ and the same speed it left with.
5Perpendicular vectors add by Pythagoras, $R = A^{2} + B^{2}$ , with the angle from $tan^{- 1}$ .
6 $[N 3 5^{\circ} E]$ starts at north and turns toward east — the first letter is the starting axis.
7Relative velocity: $v_{object/ground} = v_{object/medium} + v_{medium/ground} .$

Drop a hammer and a feather in a vacuum and they land together. That result, counter-intuitive until you have seen it, is the whole content of free fall: with air resistance removed, every object near the Earth's surface accelerates downward at the same rate, regardless of its mass. That rate is $g = 9.81 m/s^{2}$ .

Fig 1Equal time intervals, growing gaps. The spacing between images grows because the velocity keeps increasing.

Free fall is just kinematics with the acceleration filled in

There are no new equations. Free fall uses the same four kinematic equations with

a = g = 9.81 m/s^{2}

downward. Take up as positive and it becomes

a = - 9.81 m/s^{2}

; take down as positive and it becomes

+ 9.81 m/s^{2}

. Either is fine. Declaring which one you chose is not optional.

This course ignores air resistance unless a question explicitly raises it. Real falling objects do meet air resistance, which grows with speed until it balances gravity and the object reaches a constant terminal velocity — that is why a skydiver's speed levels off and why a feather loses to a hammer outside a vacuum. But in every calculation here, free fall means gravity alone.

Objects thrown upward

Throwing a ball straight up is still free fall the entire time, including on the way up. Gravity does not switch off at the top and switch back on afterwards — it acts downward throughout, which is why the ball slows on the way up, stops for an instant, and then speeds up on the way down.

At the highest point the velocity is zero, but the acceleration is not. It is still $9.81 m/s^{2}$ downward. This is the most commonly misread moment in the whole unit.
The trip is symmetric. Time up equals time down, and the speed on return to the launch height equals the launch speed — with the direction reversed.
'Find the maximum height' is a disguised instruction. It tells you $v_{f} = 0$ , which is usually the third value you needed.
Returning to the starting height means zero displacement. Over the whole flight $Δ d = 0$ — not twice the maximum height.

Adding vectors in two dimensions

Once motion leaves a straight line, signs are no longer enough and directions have to be handled properly. Two vectors are added tip to tail: draw the first, start the second where the first ended, and the resultant runs from the very start to the very end.

Fig 2Tip to tail. The resultant closes the path from the original start to the final end.

When the two vectors are perpendicular, which is the only case this course requires you to calculate, the triangle is right-angled and two tools from MCR3U finish the job: Pythagoras for the magnitude, and $tan θ$ for the angle.

R = A^{2} + B^{2} θ = tan^{- 1} (\frac{opposite}{adjacent})

Right-angled vector addition. Angles are in degrees throughout this course.

Fig 3A direction is written by naming the axis you start from, the angle you turn through, and the axis you turn toward.

Writing a direction properly

[N 3 5^{\circ} E]

means: start facing north, then rotate

3 5^{\circ}

toward the east.

[E 3 5^{\circ} N]

is a different direction — start facing east and rotate

3 5^{\circ}

toward the north. The first letter is where you start, the last is where you turn toward. Getting them backwards is a full-mark error that takes two seconds to avoid.

Relative velocity

A velocity is always measured relative to something. A passenger walking forward at $1.5 m/s$ along a train moving at $25 m/s$ is doing $1.5 m/s$ relative to the train and $26.5 m/s$ relative to the ground. Both numbers are correct; they answer different questions.

Fig 4A boat aimed straight across a flowing river does not travel straight across.

The two standard cases are a boat crossing a river and a plane flying in wind. In both, the object's velocity relative to the medium adds as a vector to the medium's velocity relative to the ground, giving the velocity relative to the ground. A boat aimed due north across an eastward current ends up travelling north-east, and the crossing time depends only on the northward part — the current sweeps it downstream but does not slow the crossing.

v_{object/ground} = v_{object/medium} + v_{medium/ground}

The inner subscripts match and cancel, leaving object-over-ground. Use that as a check that you set the sum up the right way.

Not projectile motion

Two-dimensional motion in this course means adding displacement and velocity vectors, and relative-velocity problems. Full projectile motion — an object launched at an angle, with horizontal and vertical motion analysed separately — is Grade 12 material and is not examined here.

Worked examples

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

Example 1

A stone is dropped from rest off a bridge and hits the water

2.4 s

later. How high is the bridge, and how fast is the stone moving on impact?

Example 2

A ball is thrown straight up at

16 m/s

. Find the maximum height it reaches and the total time until it returns to the thrower's hand.

Example 3

A hiker walks

45 m [N]

then

60 m [E]

. Find the resultant displacement.

Example 4

A boat that can do

4.0 m/s

in still water is aimed due north across a river

80 m

wide. The current flows east at

3.0 m/s

. Find the crossing time and the boat's velocity relative to the ground.

Where marks get lost

Saying the acceleration is zero at the top of a throw. The velocity is zero there; the acceleration is still $9.81 m/s^{2}$ downward.
Mixing sign conventions inside one problem — an upward initial velocity entered as positive alongside a downward $g$ also entered as positive.
Setting the displacement to twice the maximum height for a ball that returns to the thrower. It comes back to where it started, so $Δ d = 0$ .
Writing $[E 5 3^{\circ} N]$ when the working measured the angle from north. The first letter names the axis you started from.
Adding perpendicular vectors arithmetically: $45 + 60 = 105$ is a distance travelled, not a displacement.
Letting a river current change the crossing time. A perpendicular current sweeps the boat downstream without affecting how fast it crosses.
Leaving the calculator in radian mode, so every $tan^{- 1}$ comes back as a small decimal instead of a sensible angle in degrees.

Flashcards 9

g

Free fall

Acceleration at the top of a throw

Terminal velocity

+ 5 more

Practice

A ball is dropped from rest. How fast is it moving, in m/s, after falling for

3.00 s

? (Ignore air resistance.)

m/s

A ball is thrown straight up. At the instant it reaches its highest point, what are its velocity and acceleration?

An object falls from rest for

2.0 s

. How far does it fall, in metres?

A walker goes

9.0 km [N]

then

12.0 km [W]

. What is the magnitude of the resultant displacement in km?

A boat capable of

5.0 m/s

heads due north across a

100 m

river whose current runs east at

2.0 m/s

. How long does the crossing take?

What is the accepted magnitude of the acceleration due to gravity near Earth's surface, in

m/s^{2}

Which direction is described by

[W 2 5^{\circ} S]

A ball is thrown straight up at

12 m/s

. What maximum height, in metres, does it reach above the launch point?

A ball is thrown straight up and caught again at the same height. Put these moments in time order.

1Returns to the hand at the same speed it left
2Moving downward and speeding up
3Momentarily at rest at the top, still accelerating downward
4Leaves the hand at maximum upward speed
5Moving upward but slowing

Match each situation to the correct statement about it.

A.Acceleration zero, velocity constant
B.Crossing time unaffected, but swept downstream
C.Displacement zero, speed equal to launch speed
D.Velocity zero, acceleration still $9.81 m/s^{2}$ down

Object at the top of its flight

Skydiver at terminal velocity

Ball returning to its launch height

Boat aimed across a perpendicular current

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