fisica1-youn-e-freedman-exercicios-resolvidos

3.22: Substituting for t in terms of d in the expression for y
dart
gives
y
dart
⎛
= d
⎜ tanα0
−
⎝ 2v
2
0
gd
.
2
cos ⎟ ⎞
α0
⎠
Using the given values for d and α
0
to express this as a function of v
0
,
2 2
⎛ 26.62 m s ⎞
y = ( 3.00 m)
⎜0.90
−
.
2
⎟
⎝ v0
⎠
Then, a) y = 2.14m
, b) y =1.45m
, c) y = −2.29m
. In the last case, the dart was fired
with so slow a speed that it hit the ground before traveling the 3-meter horizontal
distance.
3.23: a) With v = 0 in Eq. (3.17), solving for t and substituting into Eq. (3.18) gives
y
2 2 2
2 2
v0
y v0
sin α
0 (30.0 m/s) sin 33.0°
( y − y ) = = =
2
2g
2g
2(9.80 m/s )
0
=
13.6 m
b) Rather than solving a quadratic, the above height may be used to find the time the
rock takes to fall from its greatest height to the ground, and hence the vertical component
of velocity,
v y
= 2yg
=
2
2(28.6 m)(9.80 m/s ) = 23.7 m/s , and so the speed of the
2
2
rock is (23.7 m/s) + ((30.0 m/s)(cos33.0°
)) = 34.6 m/s .
c) The time the rock is in the air is given by the change in the vertical component of
velocity divided by the acceleration –g; the distance is the constant horizontal
component of velocity multiplied by this time, or
( −23.7 m/s − ((30.0 m/s)sin33.0°
))
x = ( 30.0 m/s)cos33.0°
= 103 m.
2
( −9.80 m/s )
d)