Examples in Chapter 3
Problem 3.23

A man stands on the roof of a 150 m tall
building and throws a rock with a velocity of
30 m/s at an angle of 330 above the
horizontal. Ignore air resistance. Calculate:
a)
b)
c)
The maximum height above the roof reached by
the rock
The magnitude of the velocity of the rock just
before it strikes the ground
The horizontal distance from the base of the
building to the point where the rock strikes the
ground.
Step 1: Draw It!
30 m/s
Height, H
330
150 m
Range, R
What do we know?

The x- and y-components of the initial velocity



Vx0=30*cos(330)=25.16 m/s
Vy0=30*sin(330)=16.33 m/s
The acceleration in the y-direction: ay=-g=-9.8
 The acceleration in the x-direction: ax=0
 The initial height of the rock, 150 m, = y0
 The initial horizontal position of the rock, 0
Motion in the x- and y-directions
1 2
x(t )  axt  vx 0t  x0
2
vx (t )  axt  vx 0
1 2
y (t )  a yt  v y 0t  y0
2
v y (t )  a y t  v y 0
and plugging in our values
x(t )  vx 0t  25.16t
v y (t )  9.8t  16.33
y (t )  4.9t 2  16.3.3t  150
At the top of the arc, vy(t)=0
0  9.8t  16.33  t  1.666s
y (t )  4.9*(1.666)  16.3.3*1.666  150
2
y (t )  163.6 m above the ground but
H=y(t)-150=13.6 m above the roof.
The magnitude of the velocity before
it strikes ground.

The rock strikes ground when y(t)=0
0  4.9t 2  16.33t  150
Use the quadratic equation:
16.33  16.332  4( 4.9)(150
t=
2(4.9)
ignore negative time solution:
t=4.111 s
v y (t )  9.8* 4.111  16.33
v y (t )  23.96 m/s
v  v y 2  vx 2  (23.96) 2  (25.16) 2  34.78 m/s
The range of the rock

The rock strikes ground after 4.111 s
0  4.9t 2  16.33t  150
Use the quadratic equation:
16.33  16.332  4( 4.9)(150
t=
2(4.9)
ignore negative time solution:
t=4.111 s
vx (t )  25.16t  25.16* 4.111  103.4 m
Problem 3.31
In a test of a “g-suit” a volunteer is rotated
in horizontal circle of radius 7.0 m.
What is the period of rotation at which
the centripetal acceleration has a
magnitude of
a) 3.0 g?
b) 10.0 g?
Step 1: Draw It!
arad=3 g
or 10 g
7.0 m
What do we know?

arad =3 g or 10 g
 R= 7.0 m
 Need to find T
2
arad
v

R
or
arad
4 R

T
Plugging and Chugging

arad =3 g or 10 g
 R= 7.0 m
 Need to find T
4 2 R
arad 
arad
arad
T2
2
4

R
2
T 
arad
2
4

(7)
2
 3g  T 
 9.399  T  3.06 s
3*9.8
2
4

(7)
2
 10 g  T 
 2.81  T  1.67 s
10*9.8
Problem 3.37
A “moving sidewalk” in an airport moves at 1 m/s
and is 35.0 m long. If a women steps on one
end and walks at 1.5 m/s relative to the
moving sidewalk, how much time does she
require to reach the opposite side if
a) She walks in the same direction as the
sidewalk is moving?
b) She walks against the motion of the
sidewalk?
Step 1: Draw It
1.5 m/s
1 m/s
1.5 m/s
1 m/s
Must find relative velocity
Call the slide walk as reference frame A,
therefore the woman’s velocity in this
frame vA is 1.5 m/s
 Call a stationary observer frame of
reference, B and the slidewalk’s velocity
in this frame is vB =1.0 m/s
 The end points are fixed in reference B
so I must adjust the woman’s velocity to
reference B

Two different velocities

If the woman and slide are in the same
direction:
Vp/B=Vp/A+VA/B
 Vp/A= velocity of woman relative to
slidewalk=1.5 m/s
 VA/B= velocity of slidewalk relative to frame
B=1.0 m/s
 Vp/B=1+1.5=2.5 m/s

Two different velocities cont’d

If the woman and slide are in the
opposite directions:
Vp/B=Vp/A+VA/B
 Vp/A= velocity of woman relative to
slidewalk=-1.5 m/s
 VA/B= velocity of slidewalk relative to frame
B=1.0 m/s
 Vp/B=1-1.5=-0.5 m/s

Finally,

In the same direction: vp/B=d/t where
d=35 m and Vp/B=2.5 m/s


t=35/2.5=14 s
In the opposite direction: vp/B=d/t where
d=35 m and Vp/B=0.5 m/s

t=35/0.5=70 s
Problem 3.58
A baseball thrown at an angle of 600 above the
horizontal strikes a building 18 m away at a
point 8 m above the point it is thrown. Ignore
air resistance.
a) Find the magnitude of the initial velocity of
the baseball ( the velocity with which the
baseball is thrown)
b) Find the magnitude and direction of the
velocity just before it strikes the building.
Step 1: Draw It!
v0 m/s
8m
600
18 m
The Secret Weapon: The Trajectory
Equation

You can go through a lot of rigmarole but the
most powerful tool in your projectile arsenal
is this little formula below
g
2
y  tan( ) x  2
x
2
2v0 cos 
You know

x=18 m
 y= 8 m
  = 600
 You need to find v0
9.8
2
8  tan(60 )(18)  2
(18)
2
0
2v0 cos 60
0
6350
8  31.7  2  v0  16.55m / s
v0
Part B)






x=18 m, x0=0
y= 8 m, y0=0
 = 600
v0=16.55 m/s and
v0x=16.55*cos(600)
=8.275,
voy=16.55*sin(600)
=14.33
ax=0, ay=-9.8 m/s2
Need to find vx(t)
and vy(t) when
x=18 m
1 2
x(t )  axt  vx 0t  x0
2
vx (t )  axt  vx 0
1 2
y (t )  a y t  v y 0t  y0
2
v y (t )  a y t  v y 0
and plugging in our values
18  8.275t
8  4.9t 2  14.33t
18
 2.17 s
8.275
v y (t )  9.8* 2.17  14.33  6.98
t
v  vx 2  v y 2  (8.275) 2  (6.98) 2  10.8m / s
 6.98 
0
 v  tan 


40.2

 8.275 
Since y-component<0, ball is descending.
1