Thomas Calculus 13th [Solutions]

936 Chapter 13 Vector-Valued Functions and Motion in Space
23. (a)
(b)
2 2
v0 v0
2
2 2 2
R v
g
0 v0
9.8m/s
sin 2 10m = (sin 90 ) 98 m /s 9.9 m/s;
2
(9.9 m/s )
6m (sin 2 ) sin 2 0.59999 2 36.87 or 143.12 18.4 or 71.6
2
9.8m/s
24.
v 0
6
5 10 m/s and x 40cm 0.4 m; thus x ( v0
cos ) t 0.4 m
6
(5 10 m/s)(cos0 ) t
t
6 8
2
0.08 10 s 8 10 s; also, y y 1
0 ( v0 sin ) t gt
2
y
6 8 1 2 8 2 14
12
(5 10 m/s)(sin 0 )(8 10 s) (9.8m/s )(8 10 s) 3.136 10 m or 3.136 10 cm.
2
12
Therefore, it drops 3.136 10 cm.
25.
2 2
0
(400m/s)
2
v
g
R sin 2 16,000 m sin 2 sin 2 0.98 2 78.5 or 2 101.5 39.3
or 50.7
9.8m/s
26. (a)
2 2 2
0 0 0
(2 v ) 4v v
g g g
R sin 2 sin 2 4 sin or 4 times the original range.
(b) Now, let the initial range be
2 2
0 v0
( pv ) 2
g
g
2
v
R 0
sin 2 . Then we want the factor p so that pv
g
0 will double the range
sin 2 2 sin 2 p 2 p 2 or about 141%. The same percentage will
approximately double the height:
2 2
0 v0
( pv sin ) 2( sin ) 2
p
2g
2g
2 p 2.
27. The projectile reaches its maximum height when its vertical component of velocity is zero
2
dy
v v 1 v v v
v0 sin gt 0 t ymax ( v0 sin )
g
dt g g 2 g g 2g
2
( v0
sin )
2g
2 2
0 sin 0 sin 0 sin ( 0 sin ) ( 0 sin )
1
0 2
. To find the flight time we find the time when the projectile lands: ( v sin ) t gt 0
1
0 2
2v
sin
g
0
t v sin gt 0 t 0 or t . Since t 0 is the time when the projectile is fired, then
2v
sin
g
0
t is the time when the projectile strikes the ground. The range is the value of the horizontal
2 2
2v sin 2v sin v v
g 0
g g g
0 0 0 0
component when t R x ( v cos ) (2sin cos ) sin 2 . The range is
largest when 2 1 45 .
28. When marble A is located R units downrange, we have
time the height of marble A is
R
0 0 v cos
x ( v cos ) t R ( v cos ) t t . At that
1 2
y y 1
0 ( v0 sin ) t gt ( v
2 0 sin )
R g R
v cos 2 v cos
0 0
2
2
0
2
1 R
2 v cos
y R tan g . The height of marble B at the same time
2 2
0
R
v0 cos
t seconds is
1 2
1 R
2 2 v cos
2
h R tan gt R tan g . Since the heights are the same, the marbles collide regardless of
the initial velocity v 0 .
2 2
0
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