Solução_Calculo_Stewart_6e

F.
162 ¤ PROBLEMS PLUS
TX.10
Now let (a, b) be any point on or inside the parabola y = − 1
2R x2 + R 1
.Then−R ≤ a ≤ R and 0 ≤ b ≤−
2 2R a2 + R 2 .
We seek an angle α such that (a, b) lies in the path of the projectile; that is, we wish to find an angle α such that
1
b = −
2R cos 2 α a2 +(tanα) a or equivalently b = −1
2R (tan2 α +1)a 2 + (tan α) a. Rearranging this equation we get
a 2
a
2
2R tan2 α − a tan α +
2R + b =0or a 2 (tan α) 2 − 2aR(tan α)+(a 2 +2bR) =0 (∗) . This quadratic equation
for tan α has real solutions exactly when the discriminant is nonnegative. Now B 2 − 4AC ≥ 0
⇔
(−2aR) 2 − 4a 2 (a 2 +2bR) ≥ 0 ⇔ 4a 2 (R 2 − a 2 − 2bR) ≥ 0 ⇔ −a 2 − 2bR + R 2 ≥ 0 ⇔
b ≤ 1
2R (R2 − a 2 ) ⇔ b ≤ −1
2R a2 + R . This condition is satisfied since (a, b) is on or inside the parabola
2
y = − 1
2R x2 + R . It follows that (a, b) lies in the path of the projectile when tan α satisfies (∗), that is, when
2
tan α = 2aR ± 4a 2 (R 2 − a 2 − 2bR)
= R ± √ R 2 − 2bR − a 2
.
2a 2 a
(c)
If the gun is pointed at a target with height h at a distance D downrange, then
tan α = h/D. When the projectile reaches a distance D downrange (remember
we are assuming that it doesn’t hit the ground first), we have D = x =(v 0 cos α)t,
so t =
D
v 0 cos α and y =(v0 sin α)t − 1 2 gt2 = D tan α −
gD2
2v0 2 cos2 α .
Meanwhile, the target, whose x-coordinate is also D, has fallen from height h to height
h − 1 2 gt2 = D tan α −
gD2 . Thus the projectile hits the target.
2v0 2 cos2 α
5. (a) a = −g j ⇒ v = v 0 − gt j =2i − gt j ⇒ s = s 0 +2t i − 1 2 gt2 j =3.5 j +2t i − 1 2 gt2 j ⇒
s =2t i + 3.5 − 1 2 gt2 j. Therefore y =0when t = 7/g seconds. At that instant, the ball is 2 7/g ≈ 0.94 ft to the
right of the table top. Its coordinates (relative to an origin on the floor directly under the table’s edge) are (0.94, 0). At
impact, the velocity is v =2i − √ 7g j,sothespeedis|v| = √ 4+7g ≈ 15 ft/s.
7
(b) The slope of the curve when t =
g
and θ ≈ 7.6 ◦ .
dy
is
dx = dy/dt
dx/dt = −gt
2 = −g 7/g
2
= −√ √
7g
7g
.Thuscot θ =
2
2
(c) From (a), |v| = √ 4+7g. So the ball rebounds with speed 0.8 √ 4+7g ≈ 12.08 ft/s at angle of inclination
90 ◦ − θ ≈ 82.3886 ◦ . By Example 14.4.5 [ET 13.4.5], the horizontal distance traveled between bounces is
d = v2 0 sin 2α
,wherev 0 ≈ 12.08 ft/sandα ≈ 82.3886 ◦ . Therefore, d ≈ 1.197 ft. So the ball strikes the floor at
g
about 2 7/g +1.197 ≈ 2.13 ft to the right of the table’s edge.