Study Guide Holt Physics Chapter 3—Two-dimensional Motion and Vectors
I.
Section 3-1—Introduction to Vectors
A. Scalars and Vectors
1. Magnitude and Direction
a. In chapter 2 we discussed velocity. Velocity has
both a magnitude and a direction.
b. Magnitude: a measurement represented by a
number (e.g. -12 meters/second)
c. Direction: an indication of orientation. For
velocities we use a positive number for right or up
and a negative number for left or down.
2. Velocity is an example of a VECTOR. Vectors always
have both a direction and a magnitude
a. Vectors are differentiated from scalars in your
book by using BOLD type.
b. Vectors are also represented by arrows. The
longer the arrow the more magnitude the vector
has.
3. Speed is an example of a SCALAR. Scalars only have a
magnitude.
B. Properties of vectors
1. Vectors can be added graphically
a. When adding vectors, you must make sure that
they have the same units and describe similar
quantities.
b. The answer found by adding vectors together is
called the resultant.
c. Vectors can only be moved parallel to themselves in
a diagram. (see fig 3.3 on p.86)
II.
Section 3-2—Vector Operations
A. Coordinate Systems in Two Dimensions
1. We will always orient the coordinate plane so that the y-axis
lies north and south, and the x-axis lies east-west.
2. For objects flying or falling through the air we orient the yaxis vertically (up and down) and the x-axis horizontally.
B. Determining Resultant Magnitude and Direction
1. We can use right-triangle trigonometry to calculate
both the magnitude and the direction of a resultant.
2. Pythagorean Theorem: a2 + b2 = c2
3. Tangent θ = opposite/adjacent
C. Resolving Vectors Into Components
1. Vectors that are added together are called components
2. We can use the sine and cosine functions to calculate
missing sides of a right triangle.
3. Sine θ = opposite/hypotenuse
4. Cosine θ = adjacent/hypotenuse
III.
Section 3-3—Projectile Motion
A. Two-dimensional Motion
1. Projectiles are objects thrown or launched into the air.
2. Projectile motion is a form of two-dimensional motion
because the projectile has velocities along both the xand y -axes.
3. Projectiles always follow a curved path called a
“trajectory”
4. Trajectories are always shaped like a parabola*
5. Projectile motion can be thought of as free fall with
and initial horizontal velocity
6. By breaking the projectile’s trajectory into its
component vertical and horizontal component vectors
we can use right triangles (again!) to solve problems
involving trajectories like the sample problem on p.101
*in a vacuum ignoring air resistance
Projectile Motion
VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST
vf = -gΔt
vf2 = -2g∆y
Δy = -½g(∆t)2
HORIZONTAL MOTION OF A PROJECTILE
vx = vi = constant
Δx = vx Δt
Sample Problem 3D (p.101)
Given: ∆y = -321m
∆x = 45m
g = 9.81m/s2
Determine: Initial velocity (vi)
 Remember that velocity equals displacement divided by time or
v = ∆x/∆t

We know ∆x = 45 but we don’t know ∆t. Using the equations above we
can determine ∆t.

The only equation above that we can solve for ∆t is the 3rd one
∆y = -½g(Δt)2

If we substitute and solve for Δt we get
-321 = (-½)(9.81)(Δt)2
65.44 = (Δt)2
Δt = 8.1s

Now we simply divide ∆x by ∆t to get vi = 5.5m/s
B. Projectiles Launched at an Angle
1. Now let’s considered objects launched into the air at an
angle like balls and bullets.
2. Again we will break the vector up into its vertical (y)
and horizontal (x) components and use trigonometry to
make right triangles to solve problems.
vx,i = vi (cos θ)
and
vy,i = vi (sin θ)
Projectiles Launched at an Angle
PROJECTILES LAUNCHED AT AN ANGLE
vx = vi (cos θ) = constant*
Vy = Vi (sin ϴ)
Vi = √Vx2 + Vy2
ϴ = tan-1 (Vy / Vx)
*we will again ignore air resistance
Example: A projectile is launched with a velocity of 75 m/s at an angle of 600. Draw
the path of the projectile and resolve the vector into the x and y components at the
moment of launch. Calculate the maximum height, time to maximum height and
distance the projectile travels.
Given:
Vi = 75 m/s
ϴ = 600
Resolve: Vxi = Vi cos 600 = 37.5 m/s
Vyi = Vi sin 600 = 65.0 m/s
To calculate time it takes to get to top: use Vy = g x t/2
∆ t = 2Vy / g = 13.2 s
Or half the time is = Vy / g = 6.6s and total time = 13.2 s
Maximum height = ∆y = ½ g (t/2)2
Horizontal distance = ∆x = Vx (t)
t = total time = 13.2 s
∆x = (37.5 m/s) (13.2 s) = 495 m
=
215 m where t = total time.