Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

170 0 CHAPTER 13 VECTOR FUNCTIONS
23. lv(O) I = 200 m/s and, since the angle of elevation is 60°, a unit vector in the _direction of the velocity is
(cos60°)i+ (sin60°)j = ~ i + 4!j. Thusv(O) = 2oo(~i + 4j) = lOOi+ 100J3j and ifwesetuptheaxessothatthe
projectile starts at the origin, then r(O) = 0. Ignoring air resistance, the only force is that due to gravity, so
F(t) = ma(t) = -m9 j where 9 ~ 9.8 m/s 2 . Thus a(t) = -9.8j and, integrating, we have v (t) = - 9.8t j +C. But
100 i + 100 V3 j = v(O) = C , so v( t) = 100 i + ( 100 V3 - 9.8t) j and then (integrating again)
r (t) = lOOt i + (100 V3 t- 4.9t 2 ) j + D where·O = r(O) = D. Thus the position function of the projectile is
'
r(t) = lOO t i + (100 V3t- 4.9t 2 ) j .
(a) Parametric equations for the projectile are x(t) = lOOt, y(t) = 100 V3 t- 4.9t2. The projectile reaches the ground when
y(t) = O(and t > 0) => 100 V3t- 4.9t 2 = t(100 V3- 4.9t) = 0 => t = 10 4 ~f ~ 35.3 s. So the range is
x( 10 4°;0) = 100e 0 4~f3) ~ 3535 rn.
(b) The maximum height is reached when y(t) has a critical number (or equivalently, when the vertical component
of velocity is 0): y' ( t) = 0 => 100 J3 - 9.8t = 0 => t = 10 0 ~{3 ~ 17.7 s. Thus the maximum height is
v( 10~{!) = 100 V3 ( 10 ~{!) - 4.9 ( 10 9~{!r ~ 1531 m.
(c) From part (a), impact occurs at t = 10 J.;f s. Thus, the velocity at impact is
v ( 10 4°{!) = 1~0 i + [ 100 J3 - 9.8 ( 10 4°f)] j = 100 i- 100 J3j and the speed is
lv ( 10 1.f) I= JIO,OOO + 30,000 = 200 mjs.
25. As in Example 5, r (t) = (vo cos45°)t i + [(vo sin45°)t- t9t2] j = t[voJ2 t i + (voJ2 t- gt 2 ) j] . The ball lands when
y = 0 (and t > .0) ::::>:
voJ2 . . I . 1 rn2' vo J2 2 90
9 an
d
t
h
e
..
m1t1a
. I
9 g '
t = -- s. Now smce 1t ands 90 m away, 90 = x = - 2
vo v L. --or v 0 =
velocity is vo = J90g ~ 30 mjs.
27. Let a be the angle ,of elevation. Then v 0 = 150m/sand from Example 5, the horizontal distance traveled by the projectile is
d = v5 s~ 2 a. Thus 1502 ;in 2 a = 800 => sin 2a = ~~~; ~ 0.3484 => 2a ~ 20.4° or 180 - 20.4 = 159.6°.
Two angles of elevation then are a ~ 10.2° and a :=::: 79.8°.
29. Place the catapult at the origin and assume the catapult is 100 meters from the city, so the city lies between (100, 0)
and (600, 0). The initial speed is v 0 = 80 m/s and let 0 be the angle the catapult is set' at As in Example 5, the trajectory of
the catapulted rock is given byr(t) = (80cosO)ti + [(80sin0)t- 4.9t 2 ] j. The topofthe near city wall is at (100, 15),
which the rock will hit when (80 cos 0) t = 100 => t = - 4
5 O and (80 sin O)t - 4.9t
2
= 15 =>
cos
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