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Unformatted text preview: Chapter 3 Homework Solutions
P: 1, 7, 9, 11, 13
P1. A car is driven 215 km west and then 85 km southwest. What is the displacement of
the car from the point of origin (magnitude and direction)? Draw a diagram.
Solution:
N W D2x
D2y 45 ° E D1 =215 km
D2 =85 km
D North S Choose: North is "up" and "+"
East is "+"
km km
cos
sin (points west)
°
° km (points west)
km
(points south)
km
km km km km
tan km km km
km km ° (pointing south of west) P7. V is a vector 14.3 units in magnitude and points at an angle of 34.8° above the
negative x axis.
(a) Sketch this vector.
(b) Find V and V
(c) Use V and V to obtain (again) the magnitude and direction of V 1 Solution:
(a)
y 14.3
Vy 34.8 °
Vx x (b)
cos
sin °
° cos
° sin units °
units (c)
units
° tan P9. An airplane is traveling 735 km/h in a direction 41.5° west of north.
(a) Find the components of the velocity vector in the northerly and westerly directions.
(b) How far north and how far west has the plane traveled after 3.00 h?
Solution:
N
y 41.5 ° 735 km/h
W Vy
E Vx x S (a)
sin
sin °
° km/h sin km
h ° 2 cos
cos °
° (b) km/h cos , km
h
km
h km
km km
h °
!
" 3h
" 3h km
km y
By B=26. 5 A=44. 0 Ay
56.0 ° 28. 0 °
Ax Bx x C=31. 0 P11. Determine the vector A
Solution: C , given the vectors A and C in the Fig. 1 y
By B=26. 5 A=44. 0 Ay
56.0 °
Bx 28. 0 °
Ax x C=31. 0 A
C is drawn in red on the diagram. It is a vector that points from the tip of C to the
tip of A
We can determine its magnitude and direction by adding the components of A and C :
#
cos
°
#
cos
°
#
#
sin
°
#
sin
°
#
3 $
$
# $ # $ # $ # $ The magnitude of the vector then is:
%
A
$%
The direction is determined by the angle between the vector and the x axis:
#$ tan ° #$ P13. For the vectors given in the Fig. 1, determine
(a) #
&
$
'()#
&
$
' )$ # & Solution:
We first calculate the components of the three vectors:
cos
sin #
#
&
& sin cos °
°
° ° $
$
(a) # & $ # & $ # & $ # & $ # & $ %
# & ? $% tan
Note:Alternative answer for the angle:
The ** ** in the angle means that the vector is below the xaxis.
° is equavalent to
°
°
4 ° (b) # & $ # & $ # & $ # & $ # & $ %
# & ? $%
° tan
() #
&
$
Notice that #
& ?
$ # & $ this vector is going to have the same magnitude but opposite direction to the vector
from (b). Therefore:
# $ # & $ #
% &
& $ # & $ # & $% tan
°
Notice that the angle turns out to be the same as (b). The x and y components, however,
are negative meaning that the vector lies in III quadrant, while the vector in (b) has all
positive components and lies in I quadrant. Therefore, the actual angle measured from the
positive direction of the x axis is:
°
°
° 5 P: 19, 21, 31, 33, 37*, 39*
Example 38 : Level horizontal range
(a) Derive a formula for the horizontal range R of a projectile in terms of its initial
velocity ! and angle . The range means what is the horizontal distance traveled before
the projectile returns to its initial height.
(b) At what angle should it have been aimed to strike a target 320 m away.
Solution:
(a)
y
θ o=60 ° θ o=45 °
θ o=30 ° x
Range Choose:
1.
s
2.
m
3.
m
4. "up" is "+"
+
m/s
Given:
1. !
initial velocity
initial angle
Find:
R=? What is the range (how far does the water go?)
Solution:
Along y:
+
!
+
m
m!
+
!
m
s (initial time)
+
!
m
!
(the time it takes for the total flight)
+ 6 Along x:
!
m/s ,
!
!
!
! sin
cos !
+ ! ! sin
! cos !
!
sin !!
+ cos sin Note: We have used the trig. relation : sin cos sin P19.
A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s)
should the nozzle point in order that the water land 2.0 m away? Why are there two
different angles?
Solution:
We can use the formula for the range from above (it is also derived in the Textbook,
Example 38).
Given:
Range ,
m
!
m/s
Find:
?
Solution:
! sin
,
+
m/s sin
m/s
m
m/s
m/s m
sin
sin
Note: sin
cos ° ° cos
sin sin
sin cos
cos
The angle
°
°
°
Thus we have two solutions:
°
° sin 7 sin
° is also a solution P21. A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0
m from the base. What was the ball's initial speed?
Solution:
Choose:
1.
s
2.
m
3.
m
4. "up" is "+"
+
m/s
Given:
the time when it falls on the ground
')
m
(distance from the building )
3. ' )
m
(ground)
4. !
m/s
(initial velocity has no y component)
Find:
!
?
Solution:
Along y:
The y component of the final velocity is:
!
!
+
!
m/s
m/s
m
! m
s m m/s The time it took to fall on the ground then is:
!
!
+
m/s
m/s
m/s
s
We can find the initial velocity then from the position along x:
!
m
m!
s
!
!
m/s
P31. A projectile is shot from the edge of a cliff 125 m above ground level with an initial
speed of 65.0 m/s at an angle of 37.0° with the horizontal, as shown on the diagram.
(a) Determine the time taken by the projectile to hit point P at ground level
(b) Determine the range X of the projectile as measured from the base of the cliff.
At the instant just before the projectile hits point P, find
(c) the horizontal and the vertical component of its velocity
(d) the magnitude of the velocity, and
(e) the angle made by the velocity vector with the horizontal
(f) Find the maximum height above the cliff top reached by the projectile
Solution: 8 y
v o=65.0 m/s 37.0 ° 125 m x Choose: s
m
m
"up" is positive
+ Given: m/s
° !
!
!
!
! (a) m/s cos
sin °
° m/s cos
m/s sin !
! °
° m/s
m/s ? for the projectile to hit the ground
+ !
m m
m/s
m/s . m/s  m/s .
9.8 m/s 255 s m/s m/s
m/s
m/s m
m/s m
m m/s (the negative value has no physical meaning) (b) Determine the range X of the projectile as measured from the base of the cliff.
! 9 41.084 m /
m
m/s
255 s
(c) ! !
? just before it hits the ground
!
!
m/s
!
!
+
m/s
m/s
(d) ! ? just before it hits the ground
! ! 255 s m/s ! 3 01 m/s 3.07 m/s
679 m/s (e) ? just before it hits the ground
!
3.07 m/s
tan
°
!
51.9 m/s
The angle is negative because the velocity is pointing downward.
(f) 0
?
!
!
+0
!
+0
! 0 +
m/s
m/s 0 m P33. At what projection angle will the range of a projectile equal its maximum height?
Solution:
! sin
+
!
+ , 'see Example 38) 0
(see P31, part f)
In order for the range to be equal to the max. height, we need:
,
0
! sin
+ !
+ ! sin
+ remember that sin
! sin cos
+ cos sin sin cos ! sin
+ tan ° 10 ...
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This note was uploaded on 08/10/2011 for the course GOV 102 taught by Professor Smithy during the Spring '11 term at CSU San Marcos.
 Spring '11
 Smithy

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