Chapter 3: Vectors and Projectile Motion

Vectors and Scalars

You might remember from math class the term vector. We define a vector as something with

both magnitude and direction. For example, 15 meters/second west is a vector. The number part, 15, is

the magnitude or the size of the vector. West is the direction of the vector. Based on this definition,

you should now recognize three of the ideas previously studied as vectors: displacement, velocity and

acceleration. Direction is important to all these quantities.

The opposite of a vector is a scalar. A scalar has magnitude only; it lacks any directional quality.

You should recognize three of our other previously studied ideas as scalars: distance traveled, speed and

time. That time has no directional property and is a scalar will be important as a link in projectile motion

equations.

Vectors can have various notations in print. You may see an arrow over the vector, such as 𝑣→,

or it may just be bolded, as in v.

Combining Vectors

Vectors can be combined in any order. For example, if in a city, you walk four blocks north then

one block east you will arrive in the same location if you had walked one block east first, followed

second by four blocks north. Vectors can only be combined if their kind and unit match. In other words,

you could not combine a displacement vector with an acceleration vector. Nor could you combine 10

feet north and 8 meters south without converting the units first.

1. Parallel

Imagine you are at the airport watching a passenger approach a moving walkway. You note the

passenger has a walking speed of 3 m/s to your right before the moving walkway and that the walkway

is moving at 2 m/s to your right. What speed will you observe for the passenger when he is on the

walkway? If you said 5 m/s right, your instinct for adding vectors is solid. In truth, to get 5 m/s right, we

combined the 3 m/s right velocity vector with the 2 m/s right velocity vector to get 5 m/s right. They

can be expressed as one large 5 m/s right velocity vector.

2. Antiparallel

At the airport, you now notice a child walking with a speed of 3 m/s approach the moving

walkway in the wrong direction. What velocity will you notice now? You should notice the child moving

at 1 m/s to the right. It’s accurate to say we subtracted 2 m/s from 3 m/s but technically, we added a -2

m/s right vector to a 3 m/s right vector. The negative sign indicates direction so that by – 2 m/s right,

we also mean 2 m/s left.

3. Perpendicular

a) Algebraically

Imagine a tourist walks 3 city blocks east and then 4 city blocks north and wanted to know how

far he is from his starting point. The difference between distance traveled and displacement is critical

here; the distance traveled is 7 blocks but his change in position, or displacement is not. Since the two

displacement vectors are perpendicular, you probably recognize from math class how to solve this

problem. You can use the Pythagorean Theorem, d2 = 32 + 42. This happens to be a 3-4-5 triangle so the

hypotenuse and therefore the displacement is 5 blocks. But this is not a complete answer because we

also need the direction, which must include the angle. Suppose you are giving directions to the tourist.

It’s not enough to say “walk 5 blocks and you will get there.” Nor is it enough to say “walk 5 blocks

northeast and you will get there.” Because we now have all three sides of the right triangle, you can use

sine, cosine or tangent to find the angle. All methods should yield 53.1°. And so the displacement of the

tourist is 5 blocks, 53.1° northeast.

b) Graphically

There is another method for solving the combination of vectors that does not require the use of

a protractor. This method is called solving graphically and requires all vectors be drawn to scale. The

tourist first walked 3 blocks east. We’ll represent this displacement vector with a line that is 3 inches

long, as seen below.

Now, draw the next piece, also using the same scale. So 4 city blocks will be 4 inches long.

Finally, connect the two pieces with a hypotenuse and measure, working backward using the same

scale.

The hypotenuse should measure almost 5 inches, which is 5 city blocks, the same answer as the

algebraic method. We must also find the angle, by use of a protractor, which also yields about 50°.

4. Vectors Not Perpendicular

a) Algebraically

Imagine the path of a typical airplane, steep just after takeoff, then leveling out. Suppose a plane travels

50 km at 35° above the horizontal before leveling out to 15° above the horizontal for another 210 km.

What is the displacement of the plane? To solve this, use the following steps:

(1) Convert all vectors into their right angle components

(2) Add like components

(3) Redraw and use the Pythagorean Theorem

In other words, the 50 kilometers at 35° above the horizontal has components—horizontal and vertical.

That is there is a distance the plane could go along the ground and then straight up that would cause it

to arrive at the same place. Because the x-axis is perpendicular to the y-axis, we have a right triangle

and can find these two sides.

50 km

dx

So the horizontal component can be found using trig. Relating the adjacent side and the hypotenuse is

the cosine function so that cos 35 = dx/50. Or, solving for dx, we have dx = 50cos35. Therefore the

horizontal component is 41.0 km. Similarly, we can find the vertical component by using sine, or dy =

50sin35. Therefore, dy = 28.7 km.

We can do the same process for the longer leg of the plane’s trip and find another horizontal

component and another vertical component. We have

dx = 210cos15 = 202.8 km

dy = 210sin15 = 54.4 km

With these four components, we can now combine those that are alike and add the two horizontal

components and add the two vertical components.

41.0 km 202.8 km

28.7 km

54.4 km

Our total horizontal displacement is 41.0 + 202.8 = 243.8 km and our total vertical displacement is 26.8 +

54.4 = 83.1 km. Because the x-axis is perpendicular to the y-axis, we can redraw a right triangle with

243.8 km as one side and 83.1 km as the other. We can use the Pythagorean Theorem to find the

hypotenuse and therefore the displacement of the plane. d2 = 243.82 + 83.12; d = 257.6 km. Don’t

forget to use sine, cosine or tangent to find the angle. All methods should yield 18.8° above the

horizontal.

b) Graphically

This sample problem is easier to solve graphically. As we did before, begin by drawing the pieces to

scale. Begin by drawing the horizontal. Its length does not matter.

Next, use your protractor to mark off an angle of 35°.

*

Now draw the 50 km piece to scale along this angle. I will use 1 inch for 50 km.

Next, draw the plane’s second path.

Finally, connect the beginning and ending point for the plane, measure, and convert back to kilometers.

Use a protractor to measure the angle. Your graphical solution should be in close agreement to your

algebraic solution.

Projectile Motion

The ideas of vectors are so we can make better sense of objects traveling in two directions, such as a

kicked football or a launched rocket. To begin, here are some basic characteristics of projectile motion if

we consider the classic path shown below.

(1) As with the displacement of the airplane before, the velocity has components that can be found as vx = vcosθ and vy = vsinθ

(2) The velocity the projectile is shot at is the same as the velocity (but opposite direction) it lands at because gravity acts equally during its flight

(3) The angle it is launched at will be the angle it lands at (4) Complimentary angles will yield the same range. For example, a projectile launched at 30°

will travel the exact same distance as one launched at 60°. The time in the air will be quite different but the range will be equal.

(5) Hinted at by (4), the optimum angle to shoot a projectile at to maximize its range is 45°. Because of air resistance, the optimum angle in the real world is about 42°.

(6) The horizontal acceleration, ax, is always equal to zero. That is, there is nothing to affect the horizontal component of its velocity. Gravity will only affect the vertical velocity. Things that may affect its horizontal velocity will not be considered in this course: air resistance, rockets, parachutes, etc.

(7) The vertical acceleration, ay, will always be 9.8 m/s2 down due to gravity.

(8) At the projectile’s maximum height, the vertical component of the velocity, vy, has become zero. The horizontal component, vx, remains constant at this point.

Equations

The equations we have previously used now need to be modified to be specific to one direction. In

other words, we will write equations not for displacement but for horizontal displacement and a

separate equation for vertical displacement.

Starting with d = vit +1/2at2, we will replace all vectors with a subscript, yielding

dy = viyt +1/2ayt2 [Eq 4-1]

Similarly, for the horizontal direction, we will have

dx = vixt +1/2axt2

but, because ax is always zero, the second term cancels out. Also, there is no need to distinguish the

horizontal velocity as initial because it’s always unchanged. Therefore, the horizontal equation reduces

to a form that looks much like v =d/t

dx = vxt [Eq 4-2]

These are the two major equations for projectile motion. However, for maximum height calculations,

the transformation of the time independent equation is useful:

vfy2 = viy

2 + 2aydy [Eq 4-3]

Note the horizontal equivalent to this equation, when applying ax = 0, just becomes vfx = vix and is not

useful.

You may also see the three above equations in the following form:

yf =yi + viyt +1/2ayt2

xf = xi +vxt

vfy2 = viy

2 = 2ayΔy

Honors: include solving for two unknowns (see below)

Types of Projectile Motion Problems

There are three main types of projectile motion problems, though the methods for solving them are

similar. For each type, we will look at a specific example.

1. Classic

In this type of problem, the projectile is shot and returns to the same height. Usually the velocity and

the angle of the launch are given. Let’s suppose a football is kicked at 40° with a velocity of 15 m/s. We

wish to find the range and maximum height of the kick. Recognize that with the equations above, 15

m/s is of little use to us—it’s not vx nor is it vy. It’s the velocity along the hypotenuse. So, your first step

should be to find the vertical and horizontal components of the velocity using sine and cosine,

respectively. Therefore,

viy = 15sin40 = 9.64 m/s

vx = 15cos40 = 11.49 m/s

Because time is a scalar, it is useful link between the two major equations of the chapter. You should

always have enough information to solve for time from either dy = viyt +1/2ayt2 or dx = vxt. Then use that

time in the other equation. In this problem, we are solving for the range of the projectile, or dx, which

means we’ll solve the dy equation first. Critical to this type of problem is to recognize that the vertical

displacement is zero. It starts and ends on the ground; there is no change in its vertical position. We

now have

dy = viyt +1/2ayt2 0 = (9.64)t + ½(-9.8)t2 Note: use -9.8 because the direction is down

Rearranging, 4.9t2 = 9.64t. Then dividing both sides by t, we have 4.9t = 9.64.

Solving for time, t = 1.97 seconds

The entire time the projectile is in flight, all 1.97 seconds, it’s horizontal velocity is constant at 11.49

m/s. Therefore, its range is found by using dx = vxt , or dx = (11.49)(1.97) = 22.6 m

We may also need to find the object’s maximum height. Two different equations can be used. We

could solve for dy using dy = viyt +1/2ayt2 but instead of using the entire flight time, use just half of it, or

0.98 seconds. In other words, you are asking the equation, “How high up is the projectile after 0.98

seconds?” to which it can only give the maximum height. Another method is to use the time

independent equation, vfy2 = viy

2 + 2aydy to solve for dy and set vfy = 0. This time you are asking the

equation, “How high up is the projectile when its vertical velocity has become zero?” This only happens

at the maximum height. Both methods are shown below

dy = viyt +1/2ayt2 dy = 9.64(0.98)+ ½(-9.8)(0.98)2 dy = 4.37 m

vfy2 = viy

2 + 2aydy 0 = (9.64)2 + 2(-9.8)dy dy = 4.37 m

2. Horizontal Only

The next type of projectile motion problem is where the object has only a horizontal component to start

with, such as a ball rolling off a table. Let’s explore the following example. A ball rolls with a velocity of

5 m/s off of a table that is 1 meter tall. How away far from the base of the table does the ball land?

There is no need to find components of the velocities because there is no angle involved. The horizontal

velocity, vx, is 5 m/s and the initial vertical velocity is 0 m/s because the ball is moving only in the

horizontal direction. The ball will begin to acquire a vertical velocity as it rolls off the table but it doesn’t

begin with one.

Consider the two major equations again, dy = viyt +1/2ayt2 and dx = vxt. The problem asks us to find dx so

we will need to solve the dy equation first to find time and then use that time in the dx equation.

Because the initial vertical velocity is zero, and dy, the height of the table is know, we have

dy = viyt +1/2ayt2 1 m = 0 + ½(9.8)t2 Note: +9.8 can be used because to match the +1 m down

t2 = 1/4.9 ; t = 0.45 seconds

Now, using dx = vxt to solve for dx, we have dx = (5 m/s)(0.45) = 2.26 m

Some problems may not involve a table, such as a baseball traveling towards home plate, but try to

recognize them as the same type of problem.

3. Return to Different Height

The final type of projectile motion problem is when a projectile is fired at an angle but does not return

to the same height. This could be shooting a daredevil into a net or firing a rocket from a clifftop. The

process is fairly similar. Let’s use the following example: a daredevil is shot out of a cannon at 30 m/s

and an angle of 40°. He is to be caught in net 40 m from the cannon. At what height should the net be

placed?

Just as with the first type of projectile problem, you should immediately resolve the velocity into its

components. So we have

vx = 30cos40 = 23.0 m/s

viy = 30sin40 = 19.3 m/s

Yet again, we will use the two major equations, dy = viyt +1/2ayt2 and dx = vxt. This time we are interested

in dy, the height the net should be placed. This suggests we can solve the dx equation for time and use

that equation in the dy equation. Since we have the horizontal distance to the net, we can write

dx = vxt 40 m = (23.0)t t = 1.74 seconds

Then, solving for dy using this time,

dy = viyt +1/2ayt2 dy = (19.3)(1.74) +1/2(-9.8)(1.74)2 Note: -9.8 to contrast +19.3 m/s upward

The vertical displacement comes out to equal 18.75 m; this is the height the net should be

placed.

Example of Solving with Two Unknowns

A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for

the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing

location.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic

equations - x, y, vix, viy, ax, ay, and t. Because horizontal and vertical information is used separately, it is a wise idea

to organized the given information in two columns - one column for horizontal information and one column for

vertical information. In this case, the following information is either given or implied in the problem statement:

Horizontal Information Vertical Information

x = ???

vix = 2.4 m/s

ax = 0 m/s/s

y = -0.60 m

viy = 0 m/s

ay = -9.8 m/s/s

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool

ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic

equations and the known information to solve for the unknown quantities. It will almost always be the case that such

a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then

one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal

and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the

selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical

motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one

to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to

determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first

vertical equation (y = viy•t +0.5•ay•t2) will allow for the determination of the time. Once the appropriate equation has

been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values,

the equation takes the form of

-0.60 m = (0 m/s)•t + 0.5•(-9.8 m/s/s)•t2

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

-0.60 m = (-4.9 m/s/s)•t2

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

0.122 s2 = t

2

By taking the square root of both sides of the equation, the time of flight can then be determined.

t = 0.350 s (rounded from 0.3499 s)

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of

the pool ball. Recall from the given information, vix = 2.4 m/s and ax = 0 m/s/s. The first horizontal equation (x = vix•t

+ 0.5•ax•t2) can then be used to solve for "x." With the equation selected, the physics problem once more becomes

transformed into an algebra problem. By substitution of known values, the equation takes the form of

x = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

x = (2.4 m/s)•(0.3499 s)

Thus,

x = 0.84 m (rounded from 0.8398 m)

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of

0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

1. Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic

equations. For convenience sake, make a table with horizontal information on one side and vertical information

on the other side.

2. Identify the unknown quantity that the problem requests you to solve for.

3. Select either a horizontal or vertical equation to solve for the time of flight of the projectile.

4. With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal

equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a

dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by

the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1

and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a

problem that doesn't fit the mold. Problem solving is not like cooking; it is not a mere matter of following a recipe.

Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis,

and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to

solving problems.

Range of Projectile Motion Most of the basic physics textbooks talk about the horizontal range of the projectile motion. It is derived using the kinematics equations:

ax = 0 vx = vix

Δx = vixt ay = -g

viy = viy - gt Δy = viyt - ½gt2

where vix = v cos ϴ viy = v sin ϴ

Suppose a projectile is thrown from the ground level, then the range is the distance between the launch point and the landing point, where the projectile hits the ground. When the projectile comes back to the ground, the vertical displacement is zero, thus we have

0= v sin ϴ t – ½gt2 Solving for t, we have

t = 0; 2v sin ϴ g

The first solution gives the time when the projectile is thrown and the second one is the time when it hits the ground. Plugging in the second solution into the displacement equation and using 2 sin ϴ cos ϴ= sin(2ϴ), we have

R = Δx(t = 2v sin ϴ / g) = v² (sin2ϴ) g

Ex: A baseball player can throw a ball at 30.0 m/s. What is the maximum horizontal range?

Solution To maximize the range, s/he must throw a ball at an angle of 45 degrees because at this angle sin 2ϴ = 1.The range is

R = v²0 g

R= 30² 9.8

R = 91.8 m