fisica1-youn-e-freedman-exercicios-resolvidos

3.15: a) Solving Eq. (3.17) for v = 0 , with v = (15.0 m s)sin 45. 0°
,
y
0 y
(15.0 m s)sin 45°
T =
9.80 m s
2
=
1.08 s.
b) Using Equations (3.20) and (3.21) gives at t
1
, ( x,
y)
= (6.18 m, 4.52 m) :
t (11.5 m,5.74 m) : ,(16.8 m, 4.52 m) .
2
, t3
c) Using Equations (3.22) and (3.23) gives at
t1, ( vx , v
y
) = (10.6 m s, 4.9 m s) : t2
, (10.6 m s,0) t3
: (10.6 m s, − 4.9 m s), for
velocities, respectively, of 11 .7 m s @ 24.8°, 10 .6 m s @ 0° and 11 .7 m s @ −24.8°.
Note that v
x
is the same for all times, and that the y-component of velocity at t 3
is
negative that at t 1
.
d) The parallel and perpendicular components of the acceleration are obtained from
v r r v r
v ( a ⋅ v)
v v a ⋅ v v v v
a|| = , a||
= , a = a − a||
.
2
⊥
v v
v
For projectile motion, a = −g ˆ v r
j,
so a ⋅ v = −gv , and the components of acceleration
2
2
2
parallel and perpendicular to the velocity are t : −4.1 m s , 8.9 m . t : 0, 9.8 m .
2
2
t
3
: 4.1 m s ,8.9 m s .
e)
y
1
s
2
s
f) At t 1 , the projectile is moving upward but slowing down; at t 2 the motion is
instantaneously horizontal, but the vertical component of velocity is decreasing; at t 3 , the
projectile is falling down and its speed is increasing. The horizontal component of
velocity is constant.