Chapter 3: Vectors & Motion in Two Dimension
Example Questions & Problems
A x  A cos 
A y  A sin 
A
A 2x  A 2y
tan  
Ay
Ax
Example 3.1
a. Determine the numerical values of the x- & y-components of each vector A, B & C.
b. Determine the components of D (Dx & Dy), the magnitude and direction of D?
c. Sketch the resultant D using the parallelogram rule and compare it to part (b).
Example 3.2
A ball is thrown horizontally from a 20-m-high building with a speed of 5.0 m/s.
a. Sketch ball’s motion diagram in detail, draw vx vs. t and vy vs. t plots.
b. How far from the base of the building does the ball hit the ground?
Example 3.3
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of
20 m/s at an angle 5.0° above the horizontal. The horizontal distance to the net is 7.0 m, and the
net is 1.0 m high. Does the ball clear the net? If so, by how much? If not, by how much does it
miss?
Example 3.4
A rifle is aimed horizontally at a target 50 m away. The bullet hits the target 2.0 cm below the
aim point.
a. What was the bullet's flight time?
b. What was the bullet's speed as it left the barrel?
Solution
Step 1: Make a drawing, define directions, and interpret the question.
Step 2,3: Find the velocity components and setup Data Tables.
We know v0 is at 0o. The components are
v 0x  v 0cosθ  v 0 ; v 0y  v 0sinθ  0
I immediately setup data tables:
∆x
50 m
vx = vx0
?
t
−
∆y
vy
vy0
t
0-0.02 m
−
0 m/s
?
So the data table indicates that the bullet's flight time is obtained from the y-data table; the
kinematics table gives
y  v 0 y t  21 gt2  t 
2y
2(0  0.02 m)

 0.064s  t
g
9.81m/s2
b. Subsituting this into the x-table, the kinematics table leads to the equation
x
50 m
x  v 0x t  v 0x 

 782 m/s  v 0
t
0.064s
x –Data
x
vx = vox
y –Data
t
y
vy
voy
t

x-Kinematics
x
vx = vox
t
Δx = v 0x t = v x t
X
X
X
y-Kinematics
y
vy
voy
t
v y = v 0y - gt
―
X
X
X
v = v - 2gΔy
X
X
X
―
Δy = v0y t - 21 gt
X
―
X
X
X
X
X
X
2
y
2
0y
2
Δy = (v y + v 0y )t
1
2