fisica1-youn-e-freedman-exercicios-resolvidos

3.50: The bird’s tangential velocity can be found from
circumference 2π
(8.00 m) 50.27 m
v
x
=
=
= = 10.05 m/s
time of rotation 5.00 s 5.00 s
Thus its velocity consists of the components v = 10.05 m/s and v = 3.00 m/s . The speed
relative to the ground is then
2
v v x
+ v
2
=
y
x
= 10.052 + 3.002 = 10.49 m/s or 10.5 m/s
(b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the
horizontal direction, toward the center of its spiral path–and has magnitude
2
2
vx
(10.05 m/s)
2
2
ac
= =
= 12.63 m/s or 12.6 m/s
r 8.00 m
(c) Using the vertical and horizontal velocity components:
− 3.00 m/s
θ = tan 1 = 16.6°
10.05 m/s
3.51: Take + y to be downward.
Use the vertical motion to find the time in the air:
2
v = , a = 9.80 m/s , y − y = 25 m, t ?
0 y
0
y
0
=
1 2
y − y0 = v0
y
t + a t gives t = 2.259 s
2 y
During this time the dart must travel 90 m horizontally, so the horizontal component of
its velocity must be
x − x0
90 m
v0 x
= = = 40 m/s
t 2.25 s
y