Thomas Calculus 13th [Solutions]

938 Chapter 13 Vector-Valued Functions and Motion in Space
2 sin ( ) 2 cos( ) tan
g
0 0
Therefore, 0 cos( ) v v
x v
dx
d
2
0
2v
2
g
sin ( )cos( ) cos ( ) tan . If x is maximized, then OR is maximized:
2
0
2v
cos 2( ) sin 2( ) tan 0 cos 2( ) sin 2( ) tan 0
g
cot 2( ) tan 0 cot 2( ) tan tan ( ) 2( ) 90 ( ) 90
1
(90 )
1
of AOR . Therefore v
2 2
0 would bisect
32. v 0 116 ft/ sec, 45 , and x ( v0
cos ) t
369 (116 cos 45 ) t t 4.50 sec;
also
y
1 2
y ( v0 sin ) t gt
2
1
2
(116sin 45 )(4.50) (32)(4.50) 45.11 ft.
It will take the ball 4.50 sec to travel 369 ft. At that time
the ball will be 45.11 ft in the air and will hit the green
past the pin.
33. (a) (Assuming that " x " is zero at the point of impact:)
(b)
r( t) ( x( t)) i ( y( t)) j ; where x( t) (35cos 27 ) t and
2
AOR for maximum range uphill.
2
y( t) 4 (35sin 27 ) t 16 t .
2 2
( v0
sin ) (35sin 27 )
v0 y which is reached at t
sin 35sin 27
max 2g
64
0.497seconds.
(c) For the time, solve
35sin 27 35sin 27 256
32
4 4 7.945 feet,
2
y 4 (35sin 27 ) t 16t 0 for t, using the quadratic formula
2
t 1.201sec. Then the range is about x(1.201) (35cos 27 )(1.201)
37.453 feet.
(d) For the time, solve
35sin 27 35sin 27 192
32
2
y 4 (35sin 27 ) t 16t 7 for t, using the quadratic formula
2
t 0.254 and 0.740 seconds. At those times the ball is about
x (0.254) (35cos 27 )(0.254) 7.921 feet and x (0.740) (35cos 27 )(0.740) 23.077 feet from the
impact point, or about 37.453 7.921 29.532 feet and 37.453 23.077 14.376 feet from the landing
spot.
(e) Yes. It changes things because the ball wont clear the net
34. x x0 ( v0 cos ) t 0 ( v0 cos 40 ) t 0.766v0t and
0
2
( ymax
7.945).
1 2 2
y y0 ( v0 sin ) t gt 6.5 ( v
2
0 sin 40 ) t 16t
v 96.383
0t t the shot lands
v
6.5 0.643 v t 16 t ; now the shot went 73.833ft 73.833 0.766 sec;
2
96.383
v
when y 0 0 6.5 (0.643)(96.383) 16 0 68.474 v 46.6ft/sec, the
2
shots initial speed
0 0
g
0
32
148,635 148,635
0
v
68.474
35. Flight time 1sec and the measure of the angle of elevation is about 64 (using a protractor) so that
2v
sin 2v
sin 64
g
32
0 0
t 1 v 17.80ft/sec. Then
0
17.80sin 64
max 2(32)
y 4.00ft and R
2
2
v 0
sin 2
g
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