Solução_Calculo_Stewart_6e

F.
SECTION 14.4 TX.10 MOTION IN SPACE: VELOCITY AND ACCELERATION ET SECTION 13.4 ¤ 155
25. As in Example 5, r(t) =(v 0 cos 45 ◦ )t i + (v 0 sin 45 ◦ )t − 1 gt2 √ √
j = 1 v0
2 2 2 t i + v0 2 t − gt
2
j . Then the ball
lands at t = v √ √
0 2
√
s. Now since it lands 90 maway,90 = 1 v 0 2
g
2 v0 2 or v0 2 =90g and the initial velocity
g
is v 0 = √ 90g ≈ 30 m/s.
27. Let α be the angle of elevation. Then v 0 = 150 m/s and from Example 5, the horizontal distance traveled by the projectile is
d = v2 0 sin 2α
. Thus 1502 sin 2α
=800 ⇒ sin 2α = 800g
g
g
150 ≈ 0.3484 ⇒ 2α ≈ 2 20.4◦ or 180 − 20.4 =159.6 ◦ .
Two angles of elevation then are α ≈ 10.2 ◦ and α ≈ 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 =80m/sandletθ be the angle the catapult is set at. As in Example 5, the trajectory of
the catapulted rock is given by r (t) =(80cosθ)t i + (80 sin θ)t − 4.9t 2 j. The top of the near city wall is at (100, 15),
which the rock will hit when (80 cos θ) t =100 ⇒ t = 5
4cosθ and (80 sin θ)t − 4.9t2 =15
⇒
80 sin θ ·
2
5
5
4cosθ − 4.9 =15 ⇒ 100 tan θ − 7.65625 sec 2 θ =15. Replacing sec 2 θ with tan 2 θ +1gives
4cosθ
7.65625 tan 2 θ − 100 tan θ +22.62625 = 0. Using the quadratic formula, we have tan θ ≈ 0.230324, 12.8309 ⇒
θ ≈ 13.0 ◦ , 85.5 ◦ .Sofor13.0 ◦