PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: projectile motion and relative velocity (PHYS 101)

This sheet does not reprove the projectile equations, they are already in your course notes and in every search result. It treats what decides the mark: which column each number belongs to, which formula stops being true the moment the landing height changes, and how a relative velocity is written so that its direction comes out right and not merely its magnitude.

Everything is done with algebra and right triangle trigonometry, because PHYS 101 is the mechanics course WITHOUT calculus. A solution that begins by differentiating the position is not a PHYS 101 solution, even when its final number is correct.

The thread of the chapter

Two motions, one clock. The horizontal column is uniform, the vertical column is a free fall, and the only quantity allowed to cross from one column to the other is the time $t$ . Almost every mark lost in this chapter is lost on a line that mixes a horizontal datum with a vertical one.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Projectile motion
Independence of motion components
Velocity components
Time of flight
Maximum height
Range on level ground
Complementary launch angles
Parabolic trajectory
Free fall acceleration
Relative velocity
Frame of reference
River crossing problem

The essentials

The two columns, and the single quantity that crosses

• Horizontal column: nothing accelerates. $v_{x} = v_{0}\cos\theta$ at every instant, and $x = v_{0x}t$ .
• Vertical column: free fall. $v_{y} = v_{0y} - gt$ , $y = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$ , $v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0})$ , with $v_{0y} = v_{0}\sin\theta$ and $g = 9.8$ m/s $^{2}$ .
• The two columns share the clock and nothing else: $t$ is the only symbol allowed to appear in both.
• Consequence that examiners test every year: the horizontal speed has no effect on the fall. A ball rolled off a $1.225$ m bench and a ball dropped from it both land after $0.50$ s.
• Consequence used constantly in problems: find $t$ in the column where the data are complete, then carry $t$ over to the other column.

Same clock, two motions: the horizontal gaps are all equal, while the vertical drops grow as

1

3

5

7

9

. The dots are never closer together in

x

, not even at the top.

Rule the page in two before writing anything. Every classic disaster of this chapter is a single line in which a horizontal length was fed into a vertical equation, and two physical columns on the paper make that line visibly wrong.

The three level ground shortcuts, and where they die

• Time to the apex: $t_{\text{up}} = \frac{v_{0}\sin\theta}{g}$ . Always valid, whatever the landing height.
• Maximum height above the launch point: $H = \frac{(v_{0}\sin\theta)^{2}}{2g}$ . Always valid.
• Time of flight $T = \frac{2v_{0}\sin\theta}{g}$ and range $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ : valid ONLY if the landing height equals the launch height.
• On level ground $R$ is largest at $\theta = 45^{\circ}$ , where $R = \frac{v_{0}^{2}}{g}$ , and two angles adding to $90^{\circ}$ share the same range.
• Impact speed from a drop of height $h$ , angle free and time free: $v^{2} = v_{0}^{2} + 2gh$ .

Ask one question before reaching for any of these: does the projectile land at the height it left? If not, the last two lines are false, and the time comes from the quadratic $y_{\text{landing}} = v_{0y}t - \frac{1}{2}gt^{2}$ with the negative root rejected.

Relative velocity, read out loud before it is computed

• The chain rule of subscripts: $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$ , the inner letters cancelling like a fraction.
• Reversing the observer reverses the vector: $\vec{v}_{B/A} = -\vec{v}_{A/B}$ .
• The sum is a VECTOR sum: perpendicular contributions of $4.0$ and $3.0$ give $5.0$ , never $7.0$ .
• River crossing: the crossing time uses only the component PERPENDICULAR to the banks, so a current parallel to the banks never changes it.
• To land directly opposite, the upstream part of the boat velocity must cancel the current: $v_{\text{boat}}\sin\theta = v_{\text{current}}$ , impossible as soon as the current is the faster.

Say the chain aloud, in words, before writing it: rain relative to car equals rain relative to ground plus ground relative to car. That habit is what gets the DIRECTION right; the magnitude usually survives a wrong subscript order, and that is precisely why the error goes unnoticed.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which line to write, decided by what the problem does NOT mention

You do not choose an equation by the theme of the problem, you choose it by the column that already holds enough data. Find the column where only one quantity is missing, solve it there, then carry the time across.

What the problem gives you	Which column	The line to write
a height and no time	vertical	$h = \frac{1}{2}gt^{2}$ , solved for $t$
Example: Bench $1.225$ m high, ball leaving horizontally: $t = \sqrt{\frac{2(1.225)}{9.8}} = 0.50$ s, and the $4.0$ m/s never appears.
a time and a horizontal speed	horizontal	$x = v_{0x}t$
Example: $v_{0x} = 4.0$ m/s during $0.50$ s gives $x = 2.0$ m.
same launch and landing height, range asked	both, through $T$	$R = \frac{v_{0}^{2}\sin 2\theta}{g}$
Example: $20$ m/s at $45^{\circ}$ : $R = \frac{400}{9.8} = 40.8$ m, the largest range this speed can reach.
a landing height different from the launch height	vertical, as a quadratic	$y_{\text{landing}} = v_{0y}t - \frac{1}{2}gt^{2}$ , negative root rejected
Example: Thrown at $15$ m/s and $30^{\circ}$ from a cliff $20$ m high: $4.9t^{2} - 7.5t - 20 = 0$ , so $t = 2.93$ s and $x = 38.0$ m.
a drop height, impact speed asked	vertical, no time needed	$v^{2} = v_{0}^{2} + 2gh$
Example: Same cliff: $v^{2} = 15^{2} + 2(9.8)(20) = 617$ , so $v = 24.8$ m/s, and the launch angle does not enter.

The last row is the cheapest check in the chapter: it gives the impact speed without the time, so it independently confirms a root of the quadratic in one line.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. The velocity at the highest point

2 marks for the question itself, and usually the range as well, since the second half is then computed as a vertical drop

What not to write

At the top the ball has stopped, so $v = 0$ and it then falls from rest.

What to write

At the top only the VERTICAL component is zero. The velocity is horizontal and equal to $v_{0}\cos\theta$ , here $20$ m/s.

Three instants of the same flight: the arrow shortens as the ball rises, but at the top it is still

20

m/s long and horizontal, not a dot.

Why: Nothing acts horizontally, so $v_{x}$ is the same at the apex as at the launch. A truly zero velocity would mean the ball hangs still in the air for an instant, and the path after it would be a straight vertical line instead of the mirror image of the rise.

2. A horizontal length fed into a vertical equation

the whole question, because every later part inherits the wrong time

What not to write

$1.225 = (4.0)t$ , so $t = 0.31$ s.

What to write

$1.225 = \frac{1}{2}(9.8)t^{2}$ , so $t = 0.50$ s, and only then $x = (4.0)(0.50) = 2.0$ m.

Why: The $1.225$ m was measured vertically and the $4.0$ m/s horizontally: they belong to different columns and cannot meet in one equation. Two ruled columns on the page make this line visibly impossible to write.

3. The symmetric time of flight used when the heights differ

6 marks out of 10, since the time, the range and the impact velocity all follow from it

What not to write

Thrown from a $20$ m cliff at $15$ m/s and $30^{\circ}$ : $T = \frac{2v_{0y}}{g} = 1.53$ s.

What to write

$4.9t^{2} - 7.5t - 20 = 0$ , so $t = 2.93$ s, nearly twice as long.

Why: $T = \frac{2v_{0}\sin\theta}{g}$ is built on the assumption that the projectile comes back to the height it left. Off a cliff it keeps falling after that, and the range goes from $19.9$ m to $38.0$ m.

4. The launch velocity used without being resolved

2 marks, and the answer is out by a factor of nearly three

What not to write

$H = \frac{v_{0}^{2}}{2g} = \frac{25^{2}}{19.6} = 31.9$ m.

What to write

$H = \frac{(v_{0}\sin\theta)^{2}}{2g} = \frac{15^{2}}{19.6} = 11.5$ m.

Why: Only the vertical component climbs. Resolving $v_{0}$ into $v_{0}\cos\theta$ and $v_{0}\sin\theta$ is the first line of every angled launch, before any equation is chosen.

5. The trajectory read as a graph against time

2 marks, and the measurement of $g$ that the photograph was taken for

What not to write

The dots of the strobe photograph crowd together near the top, so the ball slows down there.

What to write

The gaps in $x$ are all $0.600$ m every $0.100$ s: the horizontal speed is constant at $6.00$ m/s. Only the vertical gaps shrink.

Why: The picture shows the PATH, $y$ against $x$ , not $y$ against $t$ . The slope of a path is a direction, never a velocity, and confusing the two is the same error that turns a position graph into a speed graph in the previous chapter.

6. Equal ranges taken for equal flights

the full mark of any question about hang time, clearance or the shot that passes over an obstacle

What not to write

$30^{\circ}$ and $60^{\circ}$ give the same range at the same speed, so the two shots take the same time.

What to write

The ranges match because $\sin 2\theta$ does. The times go with $\sin\theta$ : at $20$ m/s, $2.04$ s against $3.53$ s, a ratio of $\sqrt{3}$ .

Why: The steep shot buys extra time and pays for it with horizontal speed, and the two effects cancel in the product $v_{0x}T$ . The cancellation is about the range only, and says nothing about the time or the height, which differ by $\sqrt{3}$ and by $3$ .

7. A river width divided by the ground speed

3 marks, since the drift computed from the wrong time is wrong too, $48$ m instead of $60$ m

What not to write

Boat at $4.0$ m/s, current $3.0$ m/s, river $80$ m wide: $t = \frac{80}{5.0} = 16$ s.

What to write

$t = \frac{80}{4.0} = 20$ s, because only the component perpendicular to the banks crosses the river.

Why: The downstream component runs parallel to the banks for ever and brings the boat no closer to the far side. The ground speed is the speed along the diagonal path, not the speed of crossing.

8. The subscripts of a relative velocity reversed

1 or 2 marks, and the loss is invisible on the magnitude

What not to write

$\vec{v}_{\text{rain}/\text{car}} = \vec{v}_{\text{rain}/\text{ground}} + \vec{v}_{\text{car}/\text{ground}}$ .

What to write

$\vec{v}_{\text{rain}/\text{car}} = \vec{v}_{\text{rain}/\text{ground}} - \vec{v}_{\text{car}/\text{ground}}$ , since $\vec{v}_{\text{ground}/\text{car}} = -\vec{v}_{\text{car}/\text{ground}}$ .

Why: Both versions give $17$ m/s for a $15$ m/s car under an $8.0$ m/s vertical rain, so a student checking only the number sees nothing. The tilt comes out towards the rear of the car instead of the front, which is the answer the question was actually asking for.

9. The range treated as proportional to the launch speed

2 marks, plus every estimate question built on the same proportionality

What not to write

Doubling the launch speed doubles the range.

What to write

$R = \frac{v_{0}^{2}\sin 2\theta}{g}$ contains $v_{0}$ SQUARED, so doubling it multiplies the range by $4$ : at $45^{\circ}$ , $40.8$ m becomes $163$ m.

Why: The extra speed acts twice, once by sending the projectile further per second and once by keeping it in the air longer. The maximum height is multiplied by $4$ for the same reason, while the time of flight, which is linear in $v_{0}$ , is merely doubled.

Which method to choose

Which projectile line to write first

Sort the problem by the geometry, not by the story. The only question that matters at the start is whether the projectile lands at the height it left.

If the launch is HORIZONTAL → $v_{0y} = 0$ : the fall alone gives $t = \sqrt{\frac{2h}{g}}$ , then $x = v_{0}t$

Example: bench $1.225$ m high at $4.0$ m/s: $t = 0.50$ s, $x = 2.0$ m

the launch speed never enters the time, which is why the dropped ball lands with the rolled one
If the launch is at an angle and the landing height is the SAME → symmetry holds: $T = \frac{2v_{0}\sin\theta}{g}$ and $R = \frac{v_{0}^{2}\sin 2\theta}{g}$

Example: $25$ m/s with $\sin\theta = 0.60$ : $T = 3.06$ s and $R = 61.2$ m

the only case where the two compact formulas may be used, and the only case where two angles adding to $90^{\circ}$ are equivalent
If the launch is at an angle and the landing height DIFFERS → no symmetry: solve $y_{\text{landing}} = v_{0y}t - \frac{1}{2}gt^{2}$ as a quadratic and reject the negative root

Example: $15$ m/s at $30^{\circ}$ from $20$ m up: $t = 2.93$ s, $x = 38.0$ m

the height and the apex formulas still work, only the time of flight and the range formulas are lost
If the question asks for a SPEED and no time is wanted → use $v^{2} = v_{0}^{2} + 2gh$ , with $h$ the height lost

Example: same cliff: $v = \sqrt{225 + 392} = 24.8$ m/s

independent of the launch angle, which makes it the fastest check on a root of the quadratic

The apex formulas, $t_{\text{up}}$ and $H$ , sit outside this tree: they only ever use the vertical column and are true in all four branches.

Which triangle to draw when a current or a wind is involved

Draw the vector triangle before any formula. The boat or plane velocity is always given RELATIVE TO the water or the air, and the current or wind is what turns it into a velocity over the ground.

Bow pointing straight across, on the dashed line: the boat still crosses the

80

m in

20

s, but it arrives

60

m downstream of the point opposite, at the end of the solid arrow.

If the bow or nose points straight at the target and the drift is accepted → the two vectors are perpendicular: crossing time $= \frac{w}{v_{\text{boat}}}$ , drift $= v_{\text{current}} \times t$

Example: $80$ m river, boat $4.0$ m/s, current $3.0$ m/s: $t = 20$ s and a drift of $60$ m

this is always the FASTEST crossing, and the most drifted one
If the arrival point is imposed, so the drift must be cancelled → the boat vector is the HYPOTENUSE: $v_{\text{boat}}\sin\theta = v_{\text{current}}$ , and the crossing uses $\sqrt{v_{\text{boat}}^{2} - v_{\text{current}}^{2}}$

Example: same river: $\theta = 48.6^{\circ}$ upstream and $t = \frac{80}{2.65} = 30.2$ s

part of the engine is spent fighting the current, so this crossing is always the slower one
If the current is FASTER than the boat → no heading lands directly opposite. The least drift comes from the tangent to the circle of possible boat velocities

Example: current $5.0$ m/s against a $4.0$ m/s boat: bow at $53.1^{\circ}$ upstream, drift $60$ m, crossing in $33.3$ s

the tangent condition gives a right triangle with hypotenuse $v_{\text{current}}$ and leg $v_{\text{boat}}$ , so no calculus is needed

The same three branches serve an aircraft in a wind, with the air replacing the water: air speed is what the instruments read, ground speed is what the map measures, and the wind is the difference between them.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a projectile problem

When to use it: The problem gives a launch velocity, or a height and a horizontal speed, and asks for a time, a distance or an impact velocity.

1 State the axes and the origin in writing: takes up as positive, origin at the launch point, so the ground sits at $y = -20$ m.
2 Resolve the launch velocity once and for all: $v_{0x} = v_{0}\cos\theta$ and $v_{0y} = v_{0}\sin\theta$ , with the numbers.
3 Rule the page into a horizontal column and a vertical column, and put each datum under its own heading.
4 Solve for $t$ in the column that is complete, showing the equation in symbols before the numbers.
5 Carry $t$ into the other column, then answer with a sentence and a unit.

Concluding sentence

Taking up as positive with the origin at the launch point, the ball is in the air for $t = 2.93$ s and lands $38.0$ m from the foot of the cliff.

The trap: The sacrificed step is always the first one. Without a written sign convention, the marker cannot tell whether your $-20$ is a choice or a slip, and a correct negative root then looks like an error.

Marking: 1 mark for the convention and the resolved components, 1 mark for the equation written in symbols, 2 marks for the algebra, 1 mark for the sentence with its unit.

Writing up a crossing with a current or a wind

When to use it: A boat, a swimmer or an aircraft moves relative to water or air, while the water or the air moves relative to the ground.

1 Name the three frames and write the chain: velocity of the boat relative to the ground equals velocity relative to the water plus velocity of the water relative to the ground.
2 Draw the triangle, marking which vector is the hypotenuse: the boat vector when the drift is accepted, the ground vector when the arrival point is imposed.
3 Project on the two axes, across and downstream, and write one equation per axis.
4 Use the ACROSS equation for the time, the DOWNSTREAM one for the drift or for the heading.
5 Give the heading as an angle measured from a named direction, never as a bare number.

Concluding sentence

Pointing the bow $48.6^{\circ}$ upstream from the across direction cancels the current exactly, and the crossing then lasts $30.2$ s.

The trap: An angle without its reference direction earns nothing: $48.6^{\circ}$ upstream from the across direction and $41.4^{\circ}$ from the bank are the same heading, and a marker cannot guess which one you meant.

Marking: 1 mark for the chain of subscripts, 1 mark for the labelled triangle, 2 marks for the two projected equations, 1 mark for the heading stated with its reference.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Does the time belong to the right column?

Look at the line that produced $t$ . Every quantity in it must have been measured in the SAME direction. A height and a horizontal speed in the same equation is a guaranteed zero on the rest of the problem.

$1.225 = (4.0)t$ mixes a vertical metre with a horizontal speed. The correct line, $1.225 = 4.9t^{2}$ , contains only vertical quantities.
Same height at both ends?

Before using $T = \frac{2v_{0}\sin\theta}{g}$ or $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ , check that the projectile lands where it started, vertically. A cliff, a table, a raised hoop or a thrower's hand at $1.8$ m all break the symmetry.

From a $20$ m cliff the symmetric formula gives $1.53$ s against a true $2.93$ s: an error of $91$ per cent, not a rounding difference.
Is the apex velocity horizontal rather than zero?

Reread any sentence about the top of the flight. The vertical component vanishes there, the horizontal one does not, and the speed at the apex is $v_{0}\cos\theta$ .

At $25$ m/s with $\cos\theta = 0.80$ , the speed at the top is $20$ m/s. Writing zero would make the fall start from rest and shorten the range by half.
Order of magnitude, in five seconds

Multiply the horizontal speed by the flight time in your head and compare with your range. Free fall also gives quick anchors: about $5$ m in the first second, about $20$ m in two seconds.

$20$ m/s for about $3$ s must give about $60$ m. A range of $306$ m means a wrong time, a range of $11$ m means the height was copied into the range.
Does the relative velocity point the right way?

Check the DIRECTION, not only the magnitude: a reversed subscript order keeps the length and flips the tilt. Ask which observer is being served, and what that observer should plainly see.

Driving through vertical rain, the windshield gets wet and the rear window stays dry. Any answer that tilts the rain backwards is wrong even though its $17$ m/s is right.

The typical problem, taken apart

A ball thrown from a balcony, the full treatment

From a balcony $12.5$ m above the street, a ball is thrown at $v_{0} = 20$ m/s at $30^{\circ}$ above the horizontal.

Find the highest point reached above the street, the time of flight, the horizontal distance from the foot of the building, and the velocity at impact.

Take $g = 9.8$ m/s $^{2}$ and ignore the air.

The launch is

12.5

m above the landing point, so the flight is NOT symmetric: the descent lasts longer than the climb and no range formula applies.

Step 1

Up positive, origin at the balcony rail, so the street is at $y = -12.5$ m. $v_{0x} = 20\cos 30^{\circ} = 17.32$ m/s, $v_{0y} = 20\sin 30^{\circ} = 10.0$ m/s.

Why

The convention is written before any equation, because the whole exercise then rests on one signed number, the $-12.5$ . Resolving the launch velocity in the same breath removes the most common slip, using $20$ where $10.0$ belongs.

Step 2

Highest point: $H = \frac{v_{0y}^{2}}{2g} = \frac{10.0^{2}}{19.6} = 5.10$ m above the rail, that is $17.6$ m above the street.

Why

The apex formula uses the vertical column only and survives the broken symmetry. Answering in both references, above the rail and above the street, is what the marker is checking: the question said above the street.

Step 3

Time of flight: $-12.5 = 10.0t - 4.9t^{2}$ , that is $4.9t^{2} - 10.0t - 12.5 = 0$ , so $t = \frac{10.0 \pm \sqrt{100 + 245}}{9.8} = \frac{10.0 \pm 18.57}{9.8}$ and $t = 2.92$ s.

Why

This is the step where $T = \frac{2v_{0y}}{g} = 2.04$ s would have been written by reflex, and it would be wrong by almost a second. The landing height differs from the launch height, so the quadratic is solved in full.

Step 4

The second root, $t = -0.87$ s, is rejected.

Why

It is where the same parabola, extended backwards in time, would have crossed street level before the throw. It is a property of the equation, not an event, and saying so in one line earns the mark that a silent rejection loses.

Step 5

Horizontal distance: $x = v_{0x}t = (17.32)(2.92) = 50.5$ m.

Why

The time crosses from the vertical column to the horizontal one, and that is the only crossing allowed in the whole problem.

Step 6

Impact velocity: $v_{y} = 10.0 - (9.8)(2.92) = -18.57$ m/s, $v_{x} = 17.32$ m/s, so $v = \sqrt{17.32^{2} + 18.57^{2}} = \sqrt{645} = 25.4$ m/s at $47.0^{\circ}$ below the horizontal.

Why

Both components are needed because a velocity is a vector. Giving only the $18.57$ m/s, the part gravity built, is the classic half answer.

Step 7

Check without the time: $v^{2} = v_{0}^{2} + 2gh = 400 + 2(9.8)(12.5) = 645$ , so $v = 25.4$ m/s.

Why

An independent route to the same number confirms the root of the quadratic in one line. If the two disagree, the error is in the time, and nowhere else.

The conclusion, written out

The ball rises to $17.6$ m above the street, stays in the air $2.92$ s, lands $50.5$ m from the foot of the building and strikes the ground at $25.4$ m/s, at $47.0^{\circ}$ below the horizontal.

The classic mistake on this problem: The classic loss on this problem is to compute the flight time as twice the time to the apex, $2.04$ s instead of $2.92$ s. It looks harmless, it is not: the range collapses from $50.5$ m to $35.3$ m and the impact velocity comes out at $20$ m/s instead of $25.4$ m/s, so three of the four answers fall with it.

Learn by heart

• Two columns, one clock: $t$ is the only symbol allowed in both.
• At the apex $v_{y} = 0$ and $v = v_{0}\cos\theta$ , never zero.
• Always true: $t_{\text{up}} = \frac{v_{0}\sin\theta}{g}$ and $H = \frac{(v_{0}\sin\theta)^{2}}{2g}$ .
• True only at equal heights: $T = \frac{2v_{0}\sin\theta}{g}$ and $R = \frac{v_{0}^{2}\sin 2\theta}{g}$ , maximal at $45^{\circ}$ .
• Two angles adding to $90^{\circ}$ share a range; the steeper one flies $\sqrt{3}$ times longer at $30^{\circ}$ against $60^{\circ}$ , and climbs three times higher.
• Impact speed, no time needed: $v^{2} = v_{0}^{2} + 2gh$ .
• Relative velocity chains through its subscripts, $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$ , and reversing them reverses the vector.
• A river is crossed by the perpendicular component alone: bow straight across is always the quickest crossing.
• Anything released from a vehicle in level flight keeps its horizontal velocity and lands directly below it.

Frequently asked questions

Why is the velocity not zero at the top of a projectile's flight?

Only the vertical part of the velocity is zero at the top, because that is exactly what stops the climb and starts the fall. Nothing acts horizontally once the object is in the air, so the horizontal part is still the one it left with, the launch speed times the cosine of the launch angle. A ball fired at twenty five metres per second at about thirty seven degrees is still moving forward at twenty metres per second at the highest point of its path.

How do I find the time of flight when the projectile lands lower than it started?

You cannot use twice the time to the apex, because that shortcut assumes the landing point is at the launch height. Write the vertical position equation with the landing height as a negative number, move everything to one side, and solve the quadratic in time. Keep the positive root only: the negative one describes where the same parabola would have been before the throw. Thrown from a twenty metre cliff at fifteen metres per second and thirty degrees, the true flight lasts about two point nine seconds, not one point five.

Does the horizontal speed of a projectile change how fast it falls?

No, and this is the heart of the chapter. The fall is decided by the vertical column alone: the initial vertical velocity, the acceleration of gravity and the height. A ball rolled off a bench at four metres per second and a ball simply dropped from the same edge at the same instant hit the floor together, after half a second for a bench of one point two metres. The rolled ball merely travels sideways while it falls.

Why do a thirty degree and a sixty degree launch land in the same place?

On level ground the range is proportional to the sine of twice the launch angle, and doubling those two angles gives sixty and one hundred and twenty degrees, which have the same sine. So any two launch angles that add up to ninety degrees share a range. They do not share anything else: the steeper shot stays in the air about one point seven times longer and climbs three times higher, which matters as soon as an obstacle stands in the way.

How long does a boat take to cross a river that has a current?

Exactly as long as it would take in still water, provided the bow points straight at the far bank. Only the component of the velocity perpendicular to the banks crosses the river, and a current running parallel to the banks leaves that component untouched. An eighty metre river crossed by a four metre per second boat takes twenty seconds either way. What the current changes is the landing point, sixty metres downstream for a three metre per second flow.

Practise it

Corrected exercises: Projectile motion and relative velocity, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: projectile motion and relative velocity (PHYS 101)

The essentials

The two columns, and the single quantity that crosses

The three level ground shortcuts, and where they die

Relative velocity, read out loud before it is computed

The rules in table form

Which line to write, decided by what the problem does NOT mention

The mistakes that cost marks

1. The velocity at the highest point

2. A horizontal length fed into a vertical equation

3. The symmetric time of flight used when the heights differ

4. The launch velocity used without being resolved

5. The trajectory read as a graph against time

6. Equal ranges taken for equal flights

7. A river width divided by the ground speed

8. The subscripts of a relative velocity reversed

9. The range treated as proportional to the launch speed

Which method to choose

Which projectile line to write first

Which triangle to draw when a current or a wind is involved

How the answer is expected to be written

Writing up a projectile problem

Writing up a crossing with a current or a wind

Check before you hand in

The typical problem, taken apart

A ball thrown from a balcony, the full treatment

Learn by heart

Frequently asked questions

Why is the velocity not zero at the top of a projectile's flight?

How do I find the time of flight when the projectile lands lower than it started?

Does the horizontal speed of a projectile change how fast it falls?

Why do a thirty degree and a sixty degree launch land in the same place?

How long does a boat take to cross a river that has a current?

Corrected exercises: Projectile motion and relative velocity, PHYS 101 at McGill

See also

Stuck on PHYS 101 projectile problems?