fisica1-youn-e-freedman-exercicios-resolvidos

3.52: a) Setting y = −h
in Eq. (3.27) (h being the stuntwoman’s initial height above the
ground) and rearranging gives
2
2
2 2v0
sin α0
cos α0
2v0x
x −
x − h = 0,
g
g
The easier thing to do here is to recognize that this can be put in the form
2
2v
2 0xv0
y 2v0
x
x − x − h = 0,
g g
the solution to which is
v
[ ] 0 x
2
x = v0
+ v0
+ 2gh
=
y y
55.5 m (south).
g
b) The graph of v x
(t)
is a horizontal line.
3.53: The distance is the horizontal speed times the time of free fall,
2y
2(90 m)
v x
= (64.0 m/s)
= 274 m.
2
g
(9.80 m/s )
3.54: In terms of the range R and the time t that the balloon is in the air, the car’s original
distance is d = R + v t . The time t can be expressed in terms of the range and the
car
R
vcar
horizontal component of velocity, t so d = R( + )
= . Using
v0
cos α
, 1
0
v0
cos α0
2
R = v0
sin 2α0
/ g and the given values yields d = 29.5 m .