fisica1-youn-e-freedman-exercicios-resolvidos

2
3.55: a) With α
0
= 45°
, Eq. (3.27) is solved for v
gx
0 x−
y
. In this case, y = −0.9 m is the
= 2
change in height. Substitution of numerical values gives
above algebraic expression for v
0
in Eq. (3.27) gives
v = 42.8
0
m/s . b) Using the
⎛ x ⎞
y = x −
⎜ (188.9 m)
188 m
⎟
⎝ ⎠
Using x = 116 m gives y = 44.1m
above the initial height, or 45.0 m above the ground,
which is 42.0 m above the fence.
2
2
3.56: The equations of motions are y = ( v0 sin α)
t −1/
2gt
and x = ( v 0
cos α)
t , assuming
the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the
minimum velocity, y = 2D
and x = 6D
. When the match goes in the wastebasket for the
maximum velocity,
y = 2D
and x = 7D
. In both cases, sin α = cos α = 2 / 2.
.
2
To reach the minimum distance: 6 = v t , and 2D
gt
equation for t gives t =
6D
2
( ) 2
1
2 6 −
2 v0
6D
2
v0
D
2 0
2 1 2
= v
2 0t
− . Solving the first
2
. Substituting this into the second equation gives
D = D g . Solving this for v
0
gives v = 0
3 gD .
2
To reach the maximum distance: 7 = v t , and 2D
gt
equation for t gives t =
7D
2
v0
1 7D
2
gives D 7D
− g( ) 2
D
2 0
. Substituting this into the second equation
2 =
2 v
. Solving this for v
0
0
gives
as expected, is larger than the previous result.
2 1 2
= v
2 0t
−
2
. Solving the first
v = 49gD
/5 3. 13 gD , which,
0
=
3.57: The range for a projectile that lands at the same height from which it was launched
is
2
v0
sin 2α
g
v0
sin α
H 2 2
2
R = , and the maximum height that it reaches is =
g
. We must find R
when
2
sin α
D
2
H = D and v = 0
6gD
. Solving the height equation for 6gD
sin α , =
g
2
6 gD sin(70.72°
)
α . Then,
g
sin = (1/3)
1/
R = , or R = 5 .6569D
= 4 2D
.
, or