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

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1. (a) The projectile reaches maximum he~ght when 0 = ~~ = ! [(Vo sina)t- ~gt2] = vo sin a - gt; that is, when
vosina .d ( . )(vosiiJ.a) 1 (vosina) 2
t = ---an y = vosma --- - -g ---
g g 2 g
2 ° 2
Vo sm a
.
Th'
1s
. h .
1 'gh .
IS t e max1mum 1e1 t attamed when
2 g .
the projectile is fired with an angle of elevation a. This maximum height is largest when a = ~ . In that case, sin a = 1
. . . vfi
and the maximum he1ght IS-.
2g
(b) Let R = vfi /g. We are asked to consider the parabola x 2 + 2Ry - R 2 = 0 which can be rewritten as y = - __!._ x 2 + !!:. .
2R 2
The points on or inside this parabola are those for which - R ~ x ~ Rand 0 ~ y ~ ;~ x 2 + ~.
When the projectile is
fired at angle of elevation a, the points ( x, y) along its path satisfy the relations x = ( v 0 cos a) t and
y = (v 0 sina)t- tyt 2 , whereO ~ t ~ (2v 0 sina)/g(as inExan1ple 13.4.5). Thus
lxl ~ lvo cos a Cvo ;ina) I= I~ sin 2al ~ I~; I = IRI. This shows that -R ~ x ~ R.
For t in the specified range, we also have y = t ( vo sin a - ~ gt) = t gt Cvo ;in a - t) ;::: 0 and
, X 9 X g 2
2 2 x 2 =
vo cos a 2 vo cos a v 0 cos a .
0 (
y = (vosma)----- - --)2
= (tana)x -
- 1 2 R) -1 2 1 2 ( R
y- ( 2Rx + 2 = 2Rcos2 ax + 2Rx + tan a) x- 2
1 2
2 R 2 x +(tan a) x. Thus
cos a
= ~ ( 1 - _1_) + (tana)x _ R = x2 (1-sec 2 a) + 2R{tana) x - R 2
2R cos 2 a 2 2R
_ -{tan 2 a) x 2 + 2R (tan a) x - R 2 ·_ :- [{tana) x - R] 2 0
- 2R - . 2R ~
We have shown that every target that can be hit by the projecti le lies on or inside the parabola y = _ __!._ x 2 + R.
2R 2
Now let (a, b) be any point on or inside the parabola y = -
2 ~ x 2 + ~· Then·-R ~a~ Rand 0 ~ b ~ -
2 ~ a2 + ~-
We seek an angle a such that (a, b) lies in the path of the projectile; that is, we wish to find an angle a such that
b = -
2 ~c~s 2 a a 2 +(tan a) a or equivalently b = ~ (tan 2 a ~1-
1)a 2 +(tan a) a. Rearranging this equation we get
;~ tan 2 a- a tan a+(;~+ b) = 0 or a 2 (tana? - 2aR(tana) + (a 2 + 2bR) = 0 (*) . This quadratic equation
for tan a has real solutions exactly when the discriminant is nonnegative. N~w B 2 - 4AC ;::: 0
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