fisica1-youn-e-freedman-exercicios-resolvidos

2.77: Let t 1 be the fall for the watermelon, and t 2 be the travel time for the sound to
return. The total time is T = t + t 2.5 s . Let y be the height of the building, then,
2
2
1
s
2.
1 2
=
y = 1 gt and y = v t There are three equations and three unknowns. Eliminate t 2 , solve
for t 1 , and use the result to find y. A quadratic results: 1 2
gt + v t − v T = 0.
If
2 2
−b
± b −4ac
at + bt + c = 0,then
t =
2a
.
2
Here, t = t , a = 1 2 g = 4.9 m s , b = v = 340 m s,and c = −v
T = −(340
m s)(2.
5s) =
1 s
s
−
Then upon substituting these values into the quadratic formula,
2
1
s 1
s
850 m
t
1
=
− (340 m s) ±
(340 m
s)
2
2(4.9 m s
− 4(4.9 m s )( −850 m)
2
)
2
( 340 m s) ± ( 363.7 m s)
−
t
1
=
2
( ) = 2.42 s . The other solution, –71.8 s has no real physical meaning.
2 4.9 m/s
1 2 1
2
2
Then, y =
2
gt1
=
2
(9.8 m s )(2.42 s) = 28.6 m.
Check: ( 28.6 m) (340 m s) = .08s,
the
time for the sound to return.
2.78: The elevators to the observation deck of the Sears Tower in Chicago move from
the ground floor to the 103 rd floor observation deck in about 70 s. Estimating a single
floor to be about 3.5 m (11.5 ft), the average speed of the elevator is ( 103 )( 3.5 m ) = 5.15 m s.
70 s
Estimating that the elevator must come to rest in the space of one floor, the acceleration
is about
2
0 −
2
( 5.15 m s)
2
2( 3.5 m) = −3.80m s .
2
2.79: a) v = 2gh
= 2(9.80 m s )(21.3 m) = 20.4 m s;
the announcer is mistaken.
b) The required speed would be
2
2
2
v
0
= v + 2g(
y − y0
) = (25 m s) + 2(9.80 m s )( −21.3m)
= 14.4 m s,
which is not possible for a leaping diver.