Physics 107 HOMEWORK ASSIGNMENT #2
Cutnell & Johnson, 7th edition
Chapter 1: Problem 50
Chapter 2: Problems 44, 54, 56
Chapter 3: Problem 38
*50 Multiple-Concept Example 9 deals with the concepts that are important in this problem. A
grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm, due west; (2) 23.0 cm,
35.0° south of west; (3) 28.0 cm, 55.0° south of east; and (4) 35.0 cm, 63.0° north of east. Find
the magnitude and direction of the resultant displacement. Express the direction with respect to
due west.
44 Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver
springs upward with an initial speed of 1.8 m/s from a 3.0-m board. (a) Find the velocity with
which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = −3.0
m (measured from the board), assuming that the downward direction is chosen as the negative
direction.] (b) What is the highest point he reaches above the water?
*54 A ball is thrown upward from the top of a 25.0-m-tall building. The ball’s initial speed is
12.0 m/s. At the same instant, a person is running on the ground at a distance of 31.0 m from the
building. What must be the average speed of the person if he is to catch the ball at the bottom of
the building?
**56 While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When
the stone has fallen 3.20 m, you throw a second stone straight down. What initial velocity must
you give the second stone if they are both to reach the ground at the same instant? Take the
downward direction to be the negative direction.
*38 Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble
is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the
ground, the marble has a velocity that makes an angle of 65° with the horizontal. From what
height above the ground was the marble thrown?
SOLUTIONS:
50. REASONING The following table shows the components of the individual displacements
and the components of the resultant. The directions due east and due north are taken as the
positive directions.
East/West
Component
Displacement
(1)
(2)
(3)
(4)
North/South
Component
–27.0 cm
0
–(23.0 cm) cos 35.0° = –18.84 cm –(23.0 cm) sin 35.0° = –13.19 cm
(28.0 cm) cos 55.0° = 16.06 cm –(28.0 cm) sin 55.0° = –22.94 cm
(35.0 cm) cos 63.0° = 15.89 cm
(35.0 cm) sin 63.0° = 31.19 cm
Resultant
–13.89 cm
–4.94 cm
SOLUTION
a. From the Pythagorean theorem, we find that the
magnitude of the resultant displacement vector is
13.89 cm
R  (13.89 cm) 2  (4.94 cm) 2  14.7 cm
R
b. The angle  is given by
  tan 1
4.94 cm I
F
G
H13.89 cm J
K

4.94 cm
19 .6  , south of west
44. REASONING AND SOLUTION
a.
v 2  v02  2ay

v   1.8 m/s   2 –9.80 m/s 2
2
  –3.0 m   7.9 m/s
The minus is chosen, since the diver is now moving down. Hence, v  7.9 m/s .
b. The diver's velocity is zero at his highest point. The position of the diver relative to the
board is
y–
v02
2a
–
1.8 m/s 2

2 –9.80 m/s 2

 0.17 m
The position above the water is 3.0 m + 0.17 m = 3.2 m .
54. REASONING AND SOLUTION
building at any time t is
The position of the ball relative to the top of the
y  v0t  12 a t 2  v0t 
1
2
 9.80 m/s2  t 2
For the particular time that the ball arrives at the bottom of the building, y = –25.0 m. The
equation becomes
4.90t2 – 12.0t – 25.0 = 0
The quadratic equation yields solutions t = 3.79 s and –1.34 s. The negative solution is
rejected as being non-physical. During time t the person has run a distance x = vt so
v
56. REASONING
x 31.0 m

 8.18 m/s
t
3.79 s
To find the initial velocity v0,2 of the second stone, we will employ
Equation 2.8, y  v0,2t2 +
1 at 2 .
2 2
In this expression t2 is the time that the second stone is in
the air, and it is equal to the time t1 that the first stone is in the air minus the time t3.20 it
takes for the first stone to fall 3.20 m:
t2  t1  t3.20
We can find t1 and t3.20 by applying Equation 2.8 to the first stone.
SOLUTION To find the initial velocity v0,2 of the second stone, we employ Equation 2.8,
y  v0,2t2  12 at2 2 . Solving this equation for v0,2 yields
v0,2 
y  12 at2 2
t2
The time t1 for the first stone to strike the ground can be obtained from Equation 2.8,
y  v0,1t1 +
1 at 2 .
2 1
Noting that v0,1 = 0 m/s since the stone is dropped from rest and solving
this equation for t1, we have
t1 
2y

a
2  15.0 m 
 1.75 s
9.80 m/s 2
(1)
Note that the stone is falling down, so its displacement is negative (y =  15.0 m). Also, its
acceleration a is that due to gravity, so a = 9.80 m/s2.
The time t3.20 for the first stone to fall 3.20 m can also be obtained from Equation 1:
t3.20 
2y
2  3.20 m 

 0.808 s
a
9.80 m/s 2
The time t2 that the second stone is in the air is
t2  t1  t3.20  1.75 s  0.808 s  0.94 s
The initial velocity of the second stone is
v0,2 
y  12 at22
 15.0 m   12  9.80 m/s2   0.94 s 2

 11 m/s
t2
0.94 s
____________________________________________________________________
38. REASONING
Since the vertical
height is asked for, we will begin with
the vertical part of the motion,
treating it separately from the
horizontal part.
The directions
upward and to the right are chosen as
the positive directions in the drawing.
The data for the vertical motion are
summarized in the following table.
Note that the initial velocity
component v0y is zero, because the
marble is thrown horizontally. The
vertical component y of the marble’s
displacement is entered in the table as
H, where H is the height we seek.
The minus sign is included, because
the marble moves downward in the
negative y direction. The vertical
component vy of the final velocity is
checked as an important variable in
the table, because we are given the
angle that the final velocity makes
v0x
vx
65
vy
H
vx
vy
y-Direction Data
y
ay
vy
v0y
H
9.80 m/s2

0 m/s
t
with respect to the horizontal. Ignoring air resistance, we apply the equations of kinematics.
With the data indicated in the table, Equation 3.6b becomes
v 2y
2
2
2
v y  v0 y  2a y y   0 m/s   2a y   H  or H  
(1)
2a y
SOLUTION To use Equation (1), we need to determine the vertical component vy of the
final velocity. We are given that the final velocity makes an angle of 65 with respect to the
horizontal, as the inset in the drawing shows. Thus, from trigonometry, it follows that
tan 65 
v y
vx
or
v y  vx tan 65
or
v y  v0 x tan 65
where the minus sign is included, because vy points downward in the negative y direction. In
the absence of air resistance, there is no acceleration in the x direction, and the horizontal
component vx of the final velocity is equal to the initial value v0x. Substituting this result
into Equation (1) gives
H 
v 2y
2a y
 v0 x tan 65

2a y
2
  15 m/s  tan 65

 53 m
2 9.80 m/s 2
2

